Concept of Independent and identically distributed random variables Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Reference request for sum of normally distributed random variablesExpectation of number of trials before success in an urn problem without replacementUsing Binomial coefficient to solve a problem with unfair coinsHow to find out the conditional probability of an event followed by itself?Calculating independent indicator random variables7 fair coins are mixed with 3 unfair coins; PMF and expected number of flips for 3 heads?Die and toss random variable probability questionDiscrete Probability: Expected value of random variableDiscrete Probability: Random Variable Independent or Dependent?Product of independent random variables following different distributions
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Concept of Independent and identically distributed random variables
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Reference request for sum of normally distributed random variablesExpectation of number of trials before success in an urn problem without replacementUsing Binomial coefficient to solve a problem with unfair coinsHow to find out the conditional probability of an event followed by itself?Calculating independent indicator random variables7 fair coins are mixed with 3 unfair coins; PMF and expected number of flips for 3 heads?Die and toss random variable probability questionDiscrete Probability: Expected value of random variableDiscrete Probability: Random Variable Independent or Dependent?Product of independent random variables following different distributions
$begingroup$
According to https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Definition
I have a doubt regarding the two conditions above. Suppose we took a multinomial distribution where we have random variables X1, X2 ,... Xn suppose these random variables are corresponding to X1 number of red balls drawn, X2 number of blue balls drawn, and X3 number of yellow balls drawn, and suppose the probability of drawing a red ball 0.2 and blue is 0.5 and yellow is 0.3.
In Wikipedia, the first condition for two variables to be i.i.d :
$F_X1(x)=F_X2(x)$ which, I think, same as saying P(X1=x)=P(X2=x) e.g. P(X1=1)=P(X2=1) ==> 0.2 = 0.5 ?? which is not true. However, as I understood from some articles, that X1,X2, and X3 are i.i.d so How will the two conditions in Wikipedia apply (and what am I missing in this example)?
ALso, I saw many people when talking about i.i.d they are not talking about the random variables but rather the events. In the same page of wikipedia it is mentioned: "A sequence of fair or unfair coin flips is i.i.d." !!!
https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Examples
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
probability probability-theory random-variables
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
$endgroup$
add a comment |
$begingroup$
According to https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Definition
I have a doubt regarding the two conditions above. Suppose we took a multinomial distribution where we have random variables X1, X2 ,... Xn suppose these random variables are corresponding to X1 number of red balls drawn, X2 number of blue balls drawn, and X3 number of yellow balls drawn, and suppose the probability of drawing a red ball 0.2 and blue is 0.5 and yellow is 0.3.
In Wikipedia, the first condition for two variables to be i.i.d :
$F_X1(x)=F_X2(x)$ which, I think, same as saying P(X1=x)=P(X2=x) e.g. P(X1=1)=P(X2=1) ==> 0.2 = 0.5 ?? which is not true. However, as I understood from some articles, that X1,X2, and X3 are i.i.d so How will the two conditions in Wikipedia apply (and what am I missing in this example)?
ALso, I saw many people when talking about i.i.d they are not talking about the random variables but rather the events. In the same page of wikipedia it is mentioned: "A sequence of fair or unfair coin flips is i.i.d." !!!
https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Examples
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
probability probability-theory random-variables
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
$endgroup$
$begingroup$
Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
$endgroup$
– Florian
Mar 27 at 18:07
$begingroup$
@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42
add a comment |
$begingroup$
According to https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Definition
I have a doubt regarding the two conditions above. Suppose we took a multinomial distribution where we have random variables X1, X2 ,... Xn suppose these random variables are corresponding to X1 number of red balls drawn, X2 number of blue balls drawn, and X3 number of yellow balls drawn, and suppose the probability of drawing a red ball 0.2 and blue is 0.5 and yellow is 0.3.
In Wikipedia, the first condition for two variables to be i.i.d :
$F_X1(x)=F_X2(x)$ which, I think, same as saying P(X1=x)=P(X2=x) e.g. P(X1=1)=P(X2=1) ==> 0.2 = 0.5 ?? which is not true. However, as I understood from some articles, that X1,X2, and X3 are i.i.d so How will the two conditions in Wikipedia apply (and what am I missing in this example)?
ALso, I saw many people when talking about i.i.d they are not talking about the random variables but rather the events. In the same page of wikipedia it is mentioned: "A sequence of fair or unfair coin flips is i.i.d." !!!
https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Examples
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
probability probability-theory random-variables
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
$endgroup$
According to https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Definition
I have a doubt regarding the two conditions above. Suppose we took a multinomial distribution where we have random variables X1, X2 ,... Xn suppose these random variables are corresponding to X1 number of red balls drawn, X2 number of blue balls drawn, and X3 number of yellow balls drawn, and suppose the probability of drawing a red ball 0.2 and blue is 0.5 and yellow is 0.3.
In Wikipedia, the first condition for two variables to be i.i.d :
$F_X1(x)=F_X2(x)$ which, I think, same as saying P(X1=x)=P(X2=x) e.g. P(X1=1)=P(X2=1) ==> 0.2 = 0.5 ?? which is not true. However, as I understood from some articles, that X1,X2, and X3 are i.i.d so How will the two conditions in Wikipedia apply (and what am I missing in this example)?
ALso, I saw many people when talking about i.i.d they are not talking about the random variables but rather the events. In the same page of wikipedia it is mentioned: "A sequence of fair or unfair coin flips is i.i.d." !!!
https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables#Examples
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
probability probability-theory random-variables
probability probability-theory random-variables
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
edited Mar 28 at 5:41
Mosab Shaheen
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
asked Mar 27 at 17:29
Mosab ShaheenMosab Shaheen
1167
1167
$begingroup$
Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
$endgroup$
– Florian
Mar 27 at 18:07
$begingroup$
@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42
add a comment |
$begingroup$
Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
$endgroup$
– Florian
Mar 27 at 18:07
$begingroup$
@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42
$begingroup$
Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
$endgroup$
– Florian
Mar 27 at 18:07
$begingroup$
Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
$endgroup$
– Florian
Mar 27 at 18:07
$begingroup$
@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42
$begingroup$
@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
However as I know X1,X2, and X3 are i.i.d
No, they're not. They're not independent, and they're not identically distributed. Your example indicates these are coming from a multinomial distribution, so the sum $X_1+X_2+X_3$ is a constant $N$ - and that's enough to say they're definitely not independent. Similarly, with different probabilities associated to each color, they're not identically distributed.
Now, there is a family of independent identically distributed random variables associated with this. Let $Y_i$, for $i$ running from $1$ to $N$, be the result of the $i$th draw in vector form; $(1,0,0)$ if that draw is red, $(0,1,0)$ if it's blue, and $(0,0,1)$ if it's yellow. The $Y_i$ are i.i.d. The $X_i$ come from the sum of the $Y_i$; $sum_i=1^n Y_i = (X_1,X_2,X_3)$.
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Because i.i.d. is an adjective applying to (sets of) random variables, and the part you covered up with the ellipsis specifies which random variable we're talking about. Using an ellipsis this way is misquoting.
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
The random variables referred to are each the results of a single flip. Did we flip heads on the sixth flip? Did we flip heads on the third flip? The number of heads and the number of tails are not independent - and if the coin isn't fair, they're not identically distributed either.
answered Mar 27 at 18:09
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
add a comment |
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1 Answer
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$begingroup$
However as I know X1,X2, and X3 are i.i.d
No, they're not. They're not independent, and they're not identically distributed. Your example indicates these are coming from a multinomial distribution, so the sum $X_1+X_2+X_3$ is a constant $N$ - and that's enough to say they're definitely not independent. Similarly, with different probabilities associated to each color, they're not identically distributed.
Now, there is a family of independent identically distributed random variables associated with this. Let $Y_i$, for $i$ running from $1$ to $N$, be the result of the $i$th draw in vector form; $(1,0,0)$ if that draw is red, $(0,1,0)$ if it's blue, and $(0,0,1)$ if it's yellow. The $Y_i$ are i.i.d. The $X_i$ come from the sum of the $Y_i$; $sum_i=1^n Y_i = (X_1,X_2,X_3)$.
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Because i.i.d. is an adjective applying to (sets of) random variables, and the part you covered up with the ellipsis specifies which random variable we're talking about. Using an ellipsis this way is misquoting.
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
The random variables referred to are each the results of a single flip. Did we flip heads on the sixth flip? Did we flip heads on the third flip? The number of heads and the number of tails are not independent - and if the coin isn't fair, they're not identically distributed either.
answered Mar 27 at 18:09
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
add a comment |
$begingroup$
However as I know X1,X2, and X3 are i.i.d
No, they're not. They're not independent, and they're not identically distributed. Your example indicates these are coming from a multinomial distribution, so the sum $X_1+X_2+X_3$ is a constant $N$ - and that's enough to say they're definitely not independent. Similarly, with different probabilities associated to each color, they're not identically distributed.
Now, there is a family of independent identically distributed random variables associated with this. Let $Y_i$, for $i$ running from $1$ to $N$, be the result of the $i$th draw in vector form; $(1,0,0)$ if that draw is red, $(0,1,0)$ if it's blue, and $(0,0,1)$ if it's yellow. The $Y_i$ are i.i.d. The $X_i$ come from the sum of the $Y_i$; $sum_i=1^n Y_i = (X_1,X_2,X_3)$.
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Because i.i.d. is an adjective applying to (sets of) random variables, and the part you covered up with the ellipsis specifies which random variable we're talking about. Using an ellipsis this way is misquoting.
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
The random variables referred to are each the results of a single flip. Did we flip heads on the sixth flip? Did we flip heads on the third flip? The number of heads and the number of tails are not independent - and if the coin isn't fair, they're not identically distributed either.
answered Mar 27 at 18:09
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
add a comment |
$begingroup$
However as I know X1,X2, and X3 are i.i.d
No, they're not. They're not independent, and they're not identically distributed. Your example indicates these are coming from a multinomial distribution, so the sum $X_1+X_2+X_3$ is a constant $N$ - and that's enough to say they're definitely not independent. Similarly, with different probabilities associated to each color, they're not identically distributed.
Now, there is a family of independent identically distributed random variables associated with this. Let $Y_i$, for $i$ running from $1$ to $N$, be the result of the $i$th draw in vector form; $(1,0,0)$ if that draw is red, $(0,1,0)$ if it's blue, and $(0,0,1)$ if it's yellow. The $Y_i$ are i.i.d. The $X_i$ come from the sum of the $Y_i$; $sum_i=1^n Y_i = (X_1,X_2,X_3)$.
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Because i.i.d. is an adjective applying to (sets of) random variables, and the part you covered up with the ellipsis specifies which random variable we're talking about. Using an ellipsis this way is misquoting.
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
The random variables referred to are each the results of a single flip. Did we flip heads on the sixth flip? Did we flip heads on the third flip? The number of heads and the number of tails are not independent - and if the coin isn't fair, they're not identically distributed either.
answered Mar 27 at 18:09
jmerryjmerry
17k11633
$endgroup$
However as I know X1,X2, and X3 are i.i.d
No, they're not. They're not independent, and they're not identically distributed. Your example indicates these are coming from a multinomial distribution, so the sum $X_1+X_2+X_3$ is a constant $N$ - and that's enough to say they're definitely not independent. Similarly, with different probabilities associated to each color, they're not identically distributed.
Now, there is a family of independent identically distributed random variables associated with this. Let $Y_i$, for $i$ running from $1$ to $N$, be the result of the $i$th draw in vector form; $(1,0,0)$ if that draw is red, $(0,1,0)$ if it's blue, and $(0,0,1)$ if it's yellow. The $Y_i$ are i.i.d. The $X_i$ come from the sum of the $Y_i$; $sum_i=1^n Y_i = (X_1,X_2,X_3)$.
First, how is the wikipedia article saying "A sequence .... is i.i.d" they already said that i.i.d is between random variables and it is not an adjective of a "sequence" of events?
Because i.i.d. is an adjective applying to (sets of) random variables, and the part you covered up with the ellipsis specifies which random variable we're talking about. Using an ellipsis this way is misquoting.
Second, what are the random variables the article is referring to when it says "A sequence of fair or unfair coin flips is i.i.d."? if they are referring to the variables that represent the number of heads and number of tails they how will the first condition $F_X1(x)=F_X2(x)$ e.g. P(X1=1) will equal P(X2=1) in unfair coin flips?
The random variables referred to are each the results of a single flip. Did we flip heads on the sixth flip? Did we flip heads on the third flip? The number of heads and the number of tails are not independent - and if the coin isn't fair, they're not identically distributed either.
answered Mar 27 at 18:09
jmerryjmerry
17k11633
answered Mar 27 at 18:09
jmerryjmerry
17k11633
answered Mar 27 at 18:09
jmerryjmerry
17k11633
answered Mar 27 at 18:09
jmerryjmerry
17k11633
17k11633
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
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– Mosab Shaheen
Mar 28 at 6:05
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In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
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– Mosab Shaheen
Mar 28 at 6:19
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A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
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– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
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– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
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– Mosab Shaheen
Mar 28 at 8:21
add a comment |
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
Thanks. You said "the part you covered up with the ellipsis specifies which random variable we're talking about" but still the phrase covered in the ellipsis is wrong, because it is referring to the events as I mentioned above not to the random variables. They should say ""A sequence of results of fair or unfair coin flips is i.i.d." so it is the "results" who are i.i.d not the "flips". "Using an ellipsis this way is misquoting." no it is not, because I mentioned the full quoting two times before and after, and I put the ellipsis to clarify what I mean, which is very clear from the context.
$endgroup$
– Mosab Shaheen
Mar 28 at 6:05
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
In Wikipedia the article says $F_X(x)=P(X leq x)$. In your example of random variables "Let Yi, for i running from 1 to N, be the result of the ith draw in vector form...." How will "$leq$" operator be applied on a vector $Y_i$ e.g. $P(Y_1 leq (1,0,0) )$ what does smaller than means here?
$endgroup$
– Mosab Shaheen
Mar 28 at 6:19
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
A random variable can take any sort of values. That may affect the sort of operations we can do on them - for example, finding expected values requires those values to be in something that has addition and scalar multiplication - but there's nothing wrong with a random variable that takes values like "$(1,0,0)$" or "blue". It still has a probability distribution even if we can't write down some things like a cumulative distribution function.
$endgroup$
– jmerry
Mar 28 at 6:55
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
In this case what will be the first condition in i.i.d, will it be like $F_X(x)=P(X=x)$
$endgroup$
– Mosab Shaheen
Mar 28 at 8:19
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
$begingroup$
Could you give also another example for i.i.d random variables for more clarity? e.g. i.i.d random variables related to the words in a document in some Natural Language Processing (NLP) application or any other example
$endgroup$
– Mosab Shaheen
Mar 28 at 8:21
add a comment |
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Your example about "Suppose we took a multinomial distribution" is unclear. Can you develop it further ? How many balls do you draw ? How many times ?With/without replacement ? Do you have $n=3$ ? Do you think the $X_i$ should or shoud not be i.i.d. in this case ?
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– Florian
Mar 27 at 18:07
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@Florian I already said it is a multinomial distribution that means we have n draws e.g. 10 . Multinomial distribution again is defined with replacement. I took X1=1 or you can take X1=2....or X1=10. "Do you think the Xi should ..." as I understood it should but I am not convinced.
$endgroup$
– Mosab Shaheen
Mar 28 at 5:42