Binomial Distribution and the Law of Iterated ExpectationBinomial distribution and expectationBounds for the ratio of probabilities of Poisson-Binomial DistributionSum of Product of Bernoulli Random VariablesCompute Variance of Poisson Binomal-esque variableShow that this expression follows a Binomial distributionBinom(n, p): estimate the probability that the number of successes is near the mode $np$Linearity Of Expectations in Bernoulli's trialsNumber of trials follows the binomial distributionExpected number of Failures within K trials for Binomial RVCalculation of expected value.
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Binomial Distribution and the Law of Iterated Expectation
Binomial distribution and expectationBounds for the ratio of probabilities of Poisson-Binomial DistributionSum of Product of Bernoulli Random VariablesCompute Variance of Poisson Binomal-esque variableShow that this expression follows a Binomial distributionBinom(n, p): estimate the probability that the number of successes is near the mode $np$Linearity Of Expectations in Bernoulli's trialsNumber of trials follows the binomial distributionExpected number of Failures within K trials for Binomial RVCalculation of expected value.
$begingroup$
Suppose $H_i$ are random variable following Binomial($N_i, alpha$).
Conditional on $N_i$, $H_i$ is independent from past and follows Binomial($N_i, alpha$).
$N_i+1=N_i+H_i$ and $N_1=n$ (i.e., $N_1$ is not a random variable.)
I wonder whether the following holds:
$Ebig[fracH_1+H_2+...+H_tN_1+N_2+...+N_tbig]=Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_tbig]$ where $a_j$ are i.i.d. and follow Bernoulli distribution with success probability of $alpha$.
By the law of iterated expectations
$Ebig[Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_t|N_1+N_2+...+N_tbig]big]=Ebig[fracalpha(N_1+N_2+...+N_t)N_1+N_2+...+N_tbig]=alpha$
binomial-distribution expected-value
asked Mar 15 at 2:49
user3509199user3509199
416
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $H_i$ are random variable following Binomial($N_i, alpha$).
Conditional on $N_i$, $H_i$ is independent from past and follows Binomial($N_i, alpha$).
$N_i+1=N_i+H_i$ and $N_1=n$ (i.e., $N_1$ is not a random variable.)
I wonder whether the following holds:
$Ebig[fracH_1+H_2+...+H_tN_1+N_2+...+N_tbig]=Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_tbig]$ where $a_j$ are i.i.d. and follow Bernoulli distribution with success probability of $alpha$.
By the law of iterated expectations
$Ebig[Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_t|N_1+N_2+...+N_tbig]big]=Ebig[fracalpha(N_1+N_2+...+N_t)N_1+N_2+...+N_tbig]=alpha$
binomial-distribution expected-value
asked Mar 15 at 2:49
user3509199user3509199
416
$endgroup$
$begingroup$
You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
$endgroup$
– Michael
Mar 15 at 3:49
$begingroup$
Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
$endgroup$
– user3509199
Mar 15 at 3:54
$begingroup$
Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
$endgroup$
– Michael
Mar 15 at 4:00
$begingroup$
Great! Thank you.
$endgroup$
– user3509199
Mar 15 at 4:01
$begingroup$
Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
$endgroup$
– Michael
Mar 15 at 4:15
|
show 1 more comment
$begingroup$
Suppose $H_i$ are random variable following Binomial($N_i, alpha$).
Conditional on $N_i$, $H_i$ is independent from past and follows Binomial($N_i, alpha$).
$N_i+1=N_i+H_i$ and $N_1=n$ (i.e., $N_1$ is not a random variable.)
I wonder whether the following holds:
$Ebig[fracH_1+H_2+...+H_tN_1+N_2+...+N_tbig]=Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_tbig]$ where $a_j$ are i.i.d. and follow Bernoulli distribution with success probability of $alpha$.
By the law of iterated expectations
$Ebig[Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_t|N_1+N_2+...+N_tbig]big]=Ebig[fracalpha(N_1+N_2+...+N_t)N_1+N_2+...+N_tbig]=alpha$
binomial-distribution expected-value
asked Mar 15 at 2:49
user3509199user3509199
416
$endgroup$
Suppose $H_i$ are random variable following Binomial($N_i, alpha$).
Conditional on $N_i$, $H_i$ is independent from past and follows Binomial($N_i, alpha$).
$N_i+1=N_i+H_i$ and $N_1=n$ (i.e., $N_1$ is not a random variable.)
I wonder whether the following holds:
$Ebig[fracH_1+H_2+...+H_tN_1+N_2+...+N_tbig]=Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_tbig]$ where $a_j$ are i.i.d. and follow Bernoulli distribution with success probability of $alpha$.
By the law of iterated expectations
$Ebig[Ebig[fracsum_j=1^N_1a_j+sum_j=1^N_2a_j+...+sum_j=1^N_ta_jN_1+N_2+...+N_t|N_1+N_2+...+N_tbig]big]=Ebig[fracalpha(N_1+N_2+...+N_t)N_1+N_2+...+N_tbig]=alpha$
binomial-distribution expected-value
binomial-distribution expected-value
asked Mar 15 at 2:49
user3509199user3509199
416
asked Mar 15 at 2:49
user3509199user3509199
416
edited Mar 15 at 12:59
user3509199
asked Mar 15 at 2:49
user3509199user3509199
416
asked Mar 15 at 2:49
user3509199user3509199
416
asked Mar 15 at 2:49
user3509199user3509199
416
416
$begingroup$
You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
$endgroup$
– Michael
Mar 15 at 3:49
$begingroup$
Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
$endgroup$
– user3509199
Mar 15 at 3:54
$begingroup$
Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
$endgroup$
– Michael
Mar 15 at 4:00
$begingroup$
Great! Thank you.
$endgroup$
– user3509199
Mar 15 at 4:01
$begingroup$
Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
$endgroup$
– Michael
Mar 15 at 4:15
|
show 1 more comment
$begingroup$
You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
$endgroup$
– Michael
Mar 15 at 3:49
$begingroup$
Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
$endgroup$
– user3509199
Mar 15 at 3:54
$begingroup$
Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
$endgroup$
– Michael
Mar 15 at 4:00
$begingroup$
Great! Thank you.
$endgroup$
– user3509199
Mar 15 at 4:01
$begingroup$
Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
$endgroup$
– Michael
Mar 15 at 4:15
$begingroup$
You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
$endgroup$
– Michael
Mar 15 at 3:49
$begingroup$
You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
$endgroup$
– Michael
Mar 15 at 3:49
$begingroup$
Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
$endgroup$
– user3509199
Mar 15 at 3:54
$begingroup$
Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
$endgroup$
– user3509199
Mar 15 at 3:54
$begingroup$
Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
$endgroup$
– Michael
Mar 15 at 4:00
$begingroup$
Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
$endgroup$
– Michael
Mar 15 at 4:00
$begingroup$
Great! Thank you.
$endgroup$
– user3509199
Mar 15 at 4:01
$begingroup$
Great! Thank you.
$endgroup$
– user3509199
Mar 15 at 4:01
$begingroup$
Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
$endgroup$
– Michael
Mar 15 at 4:15
$begingroup$
Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
$endgroup$
– Michael
Mar 15 at 4:15
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
No. You can consider the simple case $n=1$, $t=2$. In this case we have $N_1=1$; $H_1$ is a Bernoulli random variable with success probability $alpha$; $N_2 = 1 + H_1$. Assume $0<alpha<1$. Then:
beginalign
Eleft[fracH_1+H_2N_1+N_2right] &=Eleft[fracH_1+H_22+H_1right]\
&=Eleft[Eleft[left(fracH_1+H_22+H_1right)|H_1right]right]\
&=Eleft[fracH_1]2+H_1right]\
&=Eleft[fracH_1+alpha(1+H_1)2+H_1right]\
&=left(frac1+2alpha3right)underbraceP[H_1=1]_alpha + left(fracalpha2right)underbraceP[H_1=0]_1-alpha\
&= fracalpha(alpha+5)6 \
&neq alpha
endalign
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
$endgroup$
$begingroup$
Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
$begingroup$
Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
$endgroup$
– Michael
Mar 15 at 19:12
$begingroup$
Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
$endgroup$
– user3509199
Mar 15 at 19:15
$begingroup$
Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
$endgroup$
– Michael
Mar 15 at 19:17
|
show 3 more comments
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. You can consider the simple case $n=1$, $t=2$. In this case we have $N_1=1$; $H_1$ is a Bernoulli random variable with success probability $alpha$; $N_2 = 1 + H_1$. Assume $0<alpha<1$. Then:
beginalign
Eleft[fracH_1+H_2N_1+N_2right] &=Eleft[fracH_1+H_22+H_1right]\
&=Eleft[Eleft[left(fracH_1+H_22+H_1right)|H_1right]right]\
&=Eleft[fracH_1]2+H_1right]\
&=Eleft[fracH_1+alpha(1+H_1)2+H_1right]\
&=left(frac1+2alpha3right)underbraceP[H_1=1]_alpha + left(fracalpha2right)underbraceP[H_1=0]_1-alpha\
&= fracalpha(alpha+5)6 \
&neq alpha
endalign
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
$endgroup$
$begingroup$
Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
$begingroup$
Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
$endgroup$
– Michael
Mar 15 at 19:12
$begingroup$
Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
$endgroup$
– user3509199
Mar 15 at 19:15
$begingroup$
Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
$endgroup$
– Michael
Mar 15 at 19:17
|
show 3 more comments
$begingroup$
No. You can consider the simple case $n=1$, $t=2$. In this case we have $N_1=1$; $H_1$ is a Bernoulli random variable with success probability $alpha$; $N_2 = 1 + H_1$. Assume $0<alpha<1$. Then:
beginalign
Eleft[fracH_1+H_2N_1+N_2right] &=Eleft[fracH_1+H_22+H_1right]\
&=Eleft[Eleft[left(fracH_1+H_22+H_1right)|H_1right]right]\
&=Eleft[fracH_1]2+H_1right]\
&=Eleft[fracH_1+alpha(1+H_1)2+H_1right]\
&=left(frac1+2alpha3right)underbraceP[H_1=1]_alpha + left(fracalpha2right)underbraceP[H_1=0]_1-alpha\
&= fracalpha(alpha+5)6 \
&neq alpha
endalign
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
$endgroup$
$begingroup$
Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
$begingroup$
Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
$endgroup$
– Michael
Mar 15 at 19:12
$begingroup$
Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
$endgroup$
– user3509199
Mar 15 at 19:15
$begingroup$
Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
$endgroup$
– Michael
Mar 15 at 19:17
|
show 3 more comments
$begingroup$
No. You can consider the simple case $n=1$, $t=2$. In this case we have $N_1=1$; $H_1$ is a Bernoulli random variable with success probability $alpha$; $N_2 = 1 + H_1$. Assume $0<alpha<1$. Then:
beginalign
Eleft[fracH_1+H_2N_1+N_2right] &=Eleft[fracH_1+H_22+H_1right]\
&=Eleft[Eleft[left(fracH_1+H_22+H_1right)|H_1right]right]\
&=Eleft[fracH_1]2+H_1right]\
&=Eleft[fracH_1+alpha(1+H_1)2+H_1right]\
&=left(frac1+2alpha3right)underbraceP[H_1=1]_alpha + left(fracalpha2right)underbraceP[H_1=0]_1-alpha\
&= fracalpha(alpha+5)6 \
&neq alpha
endalign
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
$endgroup$
No. You can consider the simple case $n=1$, $t=2$. In this case we have $N_1=1$; $H_1$ is a Bernoulli random variable with success probability $alpha$; $N_2 = 1 + H_1$. Assume $0<alpha<1$. Then:
beginalign
Eleft[fracH_1+H_2N_1+N_2right] &=Eleft[fracH_1+H_22+H_1right]\
&=Eleft[Eleft[left(fracH_1+H_22+H_1right)|H_1right]right]\
&=Eleft[fracH_1]2+H_1right]\
&=Eleft[fracH_1+alpha(1+H_1)2+H_1right]\
&=left(frac1+2alpha3right)underbraceP[H_1=1]_alpha + left(fracalpha2right)underbraceP[H_1=0]_1-alpha\
&= fracalpha(alpha+5)6 \
&neq alpha
endalign
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
answered Mar 15 at 16:12
MichaelMichael
13.1k11429
13.1k11429
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Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
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– Michael
Mar 15 at 16:15
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Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
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– user3509199
Mar 15 at 17:31
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Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
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– Michael
Mar 15 at 19:12
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Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
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– user3509199
Mar 15 at 19:15
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Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
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– Michael
Mar 15 at 19:17
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show 3 more comments
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Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
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Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
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– Michael
Mar 15 at 19:12
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Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
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– user3509199
Mar 15 at 19:15
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Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
$endgroup$
– Michael
Mar 15 at 19:17
$begingroup$
Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Note that $alpha(alpha+5)/6=alpha$ if and only if $alpha=0$ or $alpha=1$.
$endgroup$
– Michael
Mar 15 at 16:15
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
$begingroup$
Thanks. Could you provide some intuition why the expected ratio is not equal to $alpha$? Intuitively, given the construction of $H_i$, the ratio seems to be $alpha$.
$endgroup$
– user3509199
Mar 15 at 17:31
$begingroup$
Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
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– Michael
Mar 15 at 19:12
$begingroup$
Your method would be correct to compute: $$Eleft[fracH_iN_1+...+N_iright] = Eleft[fracE[H_iN_1+...+N_iright]=Eleft[fracalpha N_iN_1+...+N_iright]$$ However if the denominator has information about the future (after time $i$) things are different $$ Eleft[fracH_iN_1+...+N_tright]=Eleft[fracE[H_iN_1+...+N_tright] = Eleft[fracN_i+1-N_iN_1+...+N_tright]$$
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– Michael
Mar 15 at 19:12
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Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
$endgroup$
– user3509199
Mar 15 at 19:15
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Thanks for the clarification. In the revised question, I used the different conditioning. For example, now, I conditioned on $N_1+...+N_t$.
$endgroup$
– user3509199
Mar 15 at 19:15
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Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
$endgroup$
– Michael
Mar 15 at 19:17
$begingroup$
Yes, but you also jump to the incorrect conclusion $$ E[H_i|N_1+...+N_t] = E[alpha N_i|N_1+...+N_t]$$ Knowledge of the sum $N_1+...+N_t$ gives you some information about how big the $N_i+1$ value might have been, which gives you side information about $H_i$ that skews its conditional distribution in comparison to its conditional distribution given only the value of $N_i$. It would be correct to say $E[H_i|N_i]=alpha N_i$.
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– Michael
Mar 15 at 19:17
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show 3 more comments
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You say the $H_i$ are i.i.d., but from their description they seem neither independent nor identically distributed.
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– Michael
Mar 15 at 3:49
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Thanks for your comment Michael, I modified the statement a little bit. By i.i.d., I mean that $H_i$ is conditionally independent of the past, given $N_i$. Thanks!
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– user3509199
Mar 15 at 3:54
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Your calculation is good if you assume $H_i$ is conditionally independent of the past, given $N_i$. But you should not say the $H_i$ variables are "independent and identically distributed (i.i.d.)." The $H_2$ value depends on the value of $H_1$, they are not independent. Neither are they identically distributed.
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– Michael
Mar 15 at 4:00
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Great! Thank you.
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– user3509199
Mar 15 at 4:01
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Sorry, I got distracted with the conditinonings. Your denominators include information about the future, not just the past. So, how do you claim $E[fracH_2N_1+N_2+N_3|N_2]=E[fracalpha N_2N_1 + N_2 + N_3|N_2]$. It looks like $N_3$ depends on $H_2$. So I should not have said "your calculation is good."
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– Michael
Mar 15 at 4:15