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Density function of a sum of 75 discrete random variables
Density/probability function of discrete and continuous random variablesProbability density function of random variablesRandom Variables and Density FunctionRandomized recursive functionIs squared Brownian Motion a gaussian process?Probability density function of sum of independent exponential random variablesSum of independent discrete random variablesDensity function of a Function of Random VariablesIdentifying distribution functionProbability density function of the sum of two independent Levy-distributed random variables?
$begingroup$
I need help with find the density function of $S$ if
$$S=sum_1^75X_j,$$
where $X_j=I_jB_j$, and $I_1,ldots, I_75,B_1,ldots, B_75 $ are independent,
$$Pr(I_j=1)=0.01 , Pr(I_j=0)=0.99 mbox for all j,$$
$$Pr(B_j=50)=0.7, Pr(B_j=100)=0.3 mbox for 1 leq j leq 50$$
and
$$Pr(B_j=75)=0.7, Pr(B_j=150)=0.3 mbox for 51leq j leq 75$$
I think that maybe I have to calculate convolutions, but that is a tedious job and I don't know how could I progam that in python, for example. Is there an easier way to calculate the density function of S?
Thanks for any help.
probability-theory probability-distributions
edited Mar 22 at 17:20
Cettt
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
$endgroup$
add a comment |
$begingroup$
I need help with find the density function of $S$ if
$$S=sum_1^75X_j,$$
where $X_j=I_jB_j$, and $I_1,ldots, I_75,B_1,ldots, B_75 $ are independent,
$$Pr(I_j=1)=0.01 , Pr(I_j=0)=0.99 mbox for all j,$$
$$Pr(B_j=50)=0.7, Pr(B_j=100)=0.3 mbox for 1 leq j leq 50$$
and
$$Pr(B_j=75)=0.7, Pr(B_j=150)=0.3 mbox for 51leq j leq 75$$
I think that maybe I have to calculate convolutions, but that is a tedious job and I don't know how could I progam that in python, for example. Is there an easier way to calculate the density function of S?
Thanks for any help.
probability-theory probability-distributions
edited Mar 22 at 17:20
Cettt
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
$endgroup$
$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
1
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50
add a comment |
$begingroup$
I need help with find the density function of $S$ if
$$S=sum_1^75X_j,$$
where $X_j=I_jB_j$, and $I_1,ldots, I_75,B_1,ldots, B_75 $ are independent,
$$Pr(I_j=1)=0.01 , Pr(I_j=0)=0.99 mbox for all j,$$
$$Pr(B_j=50)=0.7, Pr(B_j=100)=0.3 mbox for 1 leq j leq 50$$
and
$$Pr(B_j=75)=0.7, Pr(B_j=150)=0.3 mbox for 51leq j leq 75$$
I think that maybe I have to calculate convolutions, but that is a tedious job and I don't know how could I progam that in python, for example. Is there an easier way to calculate the density function of S?
Thanks for any help.
probability-theory probability-distributions
edited Mar 22 at 17:20
Cettt
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
$endgroup$
I need help with find the density function of $S$ if
$$S=sum_1^75X_j,$$
where $X_j=I_jB_j$, and $I_1,ldots, I_75,B_1,ldots, B_75 $ are independent,
$$Pr(I_j=1)=0.01 , Pr(I_j=0)=0.99 mbox for all j,$$
$$Pr(B_j=50)=0.7, Pr(B_j=100)=0.3 mbox for 1 leq j leq 50$$
and
$$Pr(B_j=75)=0.7, Pr(B_j=150)=0.3 mbox for 51leq j leq 75$$
I think that maybe I have to calculate convolutions, but that is a tedious job and I don't know how could I progam that in python, for example. Is there an easier way to calculate the density function of S?
Thanks for any help.
probability-theory probability-distributions
probability-theory probability-distributions
edited Mar 22 at 17:20
Cettt
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
edited Mar 22 at 17:20
Cettt
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
edited Mar 22 at 17:20
Cettt
2,010623
edited Mar 22 at 17:20
Cettt
2,010623
edited Mar 22 at 17:20
Cettt
2,010623
2,010623
asked Mar 22 at 8:59
MainnetMainnet
16610
asked Mar 22 at 8:59
MainnetMainnet
16610
asked Mar 22 at 8:59
MainnetMainnet
16610
16610
$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
1
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50
add a comment |
$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
1
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50
$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
1
1
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure what do you mean by density function in this context, because I think about the probability density function which is for continuous variables and here your outcome $X_j$ is discrete, therefore, the sum $S$ of $X_j$ for $1 leq j leq 75$ will also be discrete.
However, I can give some suggestions about the probability mass function, which is a similar concept but for discrete variables.
We can begin by looking for the probability mass function of $X_j$. I think it's better to start by analysing the problem in two parts, $1$ to $50$ and then $51$ to $75$. Then compute the possible products $I_j B_j$. In this case we will have:
For $1 leq j leq 50$:
$$
beginalign
& I_j B_j = 1 times 50 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 100 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 50 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 100 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
And for $51 leq j leq 75$:
$$
beginalign
& I_j B_j = 1 times 75 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 150 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 75 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 150 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
Can you construct the probability mass function $P(X_j=x)$ from this data?
Edit: Just for fun, here is $P(X_j=x)$:
$$
P(X_j=x)
begincases
0.990 & textrmif & x=0 \
0.007 & textrmif & x=50 wedge 1 leq j leq 50 \
0.007 & textrmif & x=75 wedge 51 leq j leq 75 \
0.003 & textrmif & x=100 wedge 1 leq j leq 50 \
0.003 & textrmif & x=150 wedge 51 leq j leq 75 \
0 & & textrmotherwise
endcases
$$
And we can split the sums in two parts. Since the distribution in each part is multinomial, we get (if I'm not mistaken):
$$
sum_j=1^50 X_j = sum_j=0^50 sum_k=0^50-j (50j + 100k) times frac50!j! cdot k! cdot (50-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
$$
sum_j=51^75 X_j = sum_j=0^25 sum_k=0^25-j (75j + 150k) times frac25!j! cdot k! cdot (25-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
$endgroup$
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
add a comment |
Your Answer
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$begingroup$
I'm not sure what do you mean by density function in this context, because I think about the probability density function which is for continuous variables and here your outcome $X_j$ is discrete, therefore, the sum $S$ of $X_j$ for $1 leq j leq 75$ will also be discrete.
However, I can give some suggestions about the probability mass function, which is a similar concept but for discrete variables.
We can begin by looking for the probability mass function of $X_j$. I think it's better to start by analysing the problem in two parts, $1$ to $50$ and then $51$ to $75$. Then compute the possible products $I_j B_j$. In this case we will have:
For $1 leq j leq 50$:
$$
beginalign
& I_j B_j = 1 times 50 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 100 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 50 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 100 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
And for $51 leq j leq 75$:
$$
beginalign
& I_j B_j = 1 times 75 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 150 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 75 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 150 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
Can you construct the probability mass function $P(X_j=x)$ from this data?
Edit: Just for fun, here is $P(X_j=x)$:
$$
P(X_j=x)
begincases
0.990 & textrmif & x=0 \
0.007 & textrmif & x=50 wedge 1 leq j leq 50 \
0.007 & textrmif & x=75 wedge 51 leq j leq 75 \
0.003 & textrmif & x=100 wedge 1 leq j leq 50 \
0.003 & textrmif & x=150 wedge 51 leq j leq 75 \
0 & & textrmotherwise
endcases
$$
And we can split the sums in two parts. Since the distribution in each part is multinomial, we get (if I'm not mistaken):
$$
sum_j=1^50 X_j = sum_j=0^50 sum_k=0^50-j (50j + 100k) times frac50!j! cdot k! cdot (50-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
$$
sum_j=51^75 X_j = sum_j=0^25 sum_k=0^25-j (75j + 150k) times frac25!j! cdot k! cdot (25-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
$endgroup$
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
add a comment |
$begingroup$
I'm not sure what do you mean by density function in this context, because I think about the probability density function which is for continuous variables and here your outcome $X_j$ is discrete, therefore, the sum $S$ of $X_j$ for $1 leq j leq 75$ will also be discrete.
However, I can give some suggestions about the probability mass function, which is a similar concept but for discrete variables.
We can begin by looking for the probability mass function of $X_j$. I think it's better to start by analysing the problem in two parts, $1$ to $50$ and then $51$ to $75$. Then compute the possible products $I_j B_j$. In this case we will have:
For $1 leq j leq 50$:
$$
beginalign
& I_j B_j = 1 times 50 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 100 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 50 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 100 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
And for $51 leq j leq 75$:
$$
beginalign
& I_j B_j = 1 times 75 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 150 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 75 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 150 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
Can you construct the probability mass function $P(X_j=x)$ from this data?
Edit: Just for fun, here is $P(X_j=x)$:
$$
P(X_j=x)
begincases
0.990 & textrmif & x=0 \
0.007 & textrmif & x=50 wedge 1 leq j leq 50 \
0.007 & textrmif & x=75 wedge 51 leq j leq 75 \
0.003 & textrmif & x=100 wedge 1 leq j leq 50 \
0.003 & textrmif & x=150 wedge 51 leq j leq 75 \
0 & & textrmotherwise
endcases
$$
And we can split the sums in two parts. Since the distribution in each part is multinomial, we get (if I'm not mistaken):
$$
sum_j=1^50 X_j = sum_j=0^50 sum_k=0^50-j (50j + 100k) times frac50!j! cdot k! cdot (50-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
$$
sum_j=51^75 X_j = sum_j=0^25 sum_k=0^25-j (75j + 150k) times frac25!j! cdot k! cdot (25-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
$endgroup$
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
add a comment |
$begingroup$
I'm not sure what do you mean by density function in this context, because I think about the probability density function which is for continuous variables and here your outcome $X_j$ is discrete, therefore, the sum $S$ of $X_j$ for $1 leq j leq 75$ will also be discrete.
However, I can give some suggestions about the probability mass function, which is a similar concept but for discrete variables.
We can begin by looking for the probability mass function of $X_j$. I think it's better to start by analysing the problem in two parts, $1$ to $50$ and then $51$ to $75$. Then compute the possible products $I_j B_j$. In this case we will have:
For $1 leq j leq 50$:
$$
beginalign
& I_j B_j = 1 times 50 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 100 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 50 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 100 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
And for $51 leq j leq 75$:
$$
beginalign
& I_j B_j = 1 times 75 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 150 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 75 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 150 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
Can you construct the probability mass function $P(X_j=x)$ from this data?
Edit: Just for fun, here is $P(X_j=x)$:
$$
P(X_j=x)
begincases
0.990 & textrmif & x=0 \
0.007 & textrmif & x=50 wedge 1 leq j leq 50 \
0.007 & textrmif & x=75 wedge 51 leq j leq 75 \
0.003 & textrmif & x=100 wedge 1 leq j leq 50 \
0.003 & textrmif & x=150 wedge 51 leq j leq 75 \
0 & & textrmotherwise
endcases
$$
And we can split the sums in two parts. Since the distribution in each part is multinomial, we get (if I'm not mistaken):
$$
sum_j=1^50 X_j = sum_j=0^50 sum_k=0^50-j (50j + 100k) times frac50!j! cdot k! cdot (50-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
$$
sum_j=51^75 X_j = sum_j=0^25 sum_k=0^25-j (75j + 150k) times frac25!j! cdot k! cdot (25-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
$endgroup$
I'm not sure what do you mean by density function in this context, because I think about the probability density function which is for continuous variables and here your outcome $X_j$ is discrete, therefore, the sum $S$ of $X_j$ for $1 leq j leq 75$ will also be discrete.
However, I can give some suggestions about the probability mass function, which is a similar concept but for discrete variables.
We can begin by looking for the probability mass function of $X_j$. I think it's better to start by analysing the problem in two parts, $1$ to $50$ and then $51$ to $75$. Then compute the possible products $I_j B_j$. In this case we will have:
For $1 leq j leq 50$:
$$
beginalign
& I_j B_j = 1 times 50 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 100 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 50 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 100 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
And for $51 leq j leq 75$:
$$
beginalign
& I_j B_j = 1 times 75 & textrmwith quad Pr = 0.01 times 0.7 \
& I_j B_j = 1 times 150 & textrmwith quad Pr = 0.01 times 0.3 \
& I_j B_j = 0 times 75 & textrmwith quad Pr = 0.99 times 0.7 \
& I_j B_j = 0 times 150 & textrmwith quad Pr = 0.99 times 0.3
endalign
$$
Can you construct the probability mass function $P(X_j=x)$ from this data?
Edit: Just for fun, here is $P(X_j=x)$:
$$
P(X_j=x)
begincases
0.990 & textrmif & x=0 \
0.007 & textrmif & x=50 wedge 1 leq j leq 50 \
0.007 & textrmif & x=75 wedge 51 leq j leq 75 \
0.003 & textrmif & x=100 wedge 1 leq j leq 50 \
0.003 & textrmif & x=150 wedge 51 leq j leq 75 \
0 & & textrmotherwise
endcases
$$
And we can split the sums in two parts. Since the distribution in each part is multinomial, we get (if I'm not mistaken):
$$
sum_j=1^50 X_j = sum_j=0^50 sum_k=0^50-j (50j + 100k) times frac50!j! cdot k! cdot (50-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
$$
sum_j=51^75 X_j = sum_j=0^25 sum_k=0^25-j (75j + 150k) times frac25!j! cdot k! cdot (25-j-k)! times 0.007^j times 0.003^k times 0.990^50-j
$$
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
edited Mar 24 at 21:04
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
answered Mar 23 at 11:48
ErtxiemErtxiem
665112
665112
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
add a comment |
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
$begingroup$
Thank you. I used python to calculate the coefficients of the probability generating function of $S$ and so get the probability $P(S=s)$.
$endgroup$
– Mainnet
Mar 24 at 14:10
1
1
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
$begingroup$
You're welcome, @Minysh. Just for fun, I added $P(X_j=x). :)
$endgroup$
– Ertxiem
Mar 24 at 20:46
add a comment |
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$begingroup$
You have to show exact form of this pdf, or just program it in python?
$endgroup$
– vermator
Mar 22 at 9:29
$begingroup$
I have to show a exact form of this pdf, but I can program it in python.
$endgroup$
– Mainnet
Mar 22 at 9:35
$begingroup$
Are $ I_j, B_j $ all indepenent?
$endgroup$
– lonza leggiera
Mar 22 at 10:54
$begingroup$
yes, they are independent
$endgroup$
– Mainnet
Mar 22 at 10:59
1
$begingroup$
@Minysh: I've added an answer that I hope that can help you. I can add more information to it, but I would need to understand better what do you need.
$endgroup$
– Ertxiem
Mar 23 at 11:50