Mixing
Conditional variance and expectancy of two independent poisson variables
$begingroup$
Le $ X sim Pois(5) , Y sim Pois(10) $ both independent. Suppose I draw and rectangle with width $ X $ and length $ Y $. Suppose the circumference of the rectangle is $ 28 $ what is $ Var(X) $ ?
First I set another variable $ S $ to be the circumference . So $ S = 2X + 2Y $
$$ Var(X) = E[Var(X|S=28)] + Var(E[X|S = 28]) = \E[Var(X|Y=14-X)] + Var(E[X|Y=14-X]) $$
I'm really having hard time grasping this concept, what is $ E[X|S = 28 ] $ ? and what is the variance of such variable? Maybe I should set new variable $ T = (X|S=28) $ and try to understand what distribution it has?
conditional-expectation poisson-distribution variance
asked Jan 18 at 13:14
bm1125bm1125
65116
$endgroup$
add a comment |
$begingroup$
Le $ X sim Pois(5) , Y sim Pois(10) $ both independent. Suppose I draw and rectangle with width $ X $ and length $ Y $. Suppose the circumference of the rectangle is $ 28 $ what is $ Var(X) $ ?
First I set another variable $ S $ to be the circumference . So $ S = 2X + 2Y $
$$ Var(X) = E[Var(X|S=28)] + Var(E[X|S = 28]) = \E[Var(X|Y=14-X)] + Var(E[X|Y=14-X]) $$
I'm really having hard time grasping this concept, what is $ E[X|S = 28 ] $ ? and what is the variance of such variable? Maybe I should set new variable $ T = (X|S=28) $ and try to understand what distribution it has?
conditional-expectation poisson-distribution variance
asked Jan 18 at 13:14
bm1125bm1125
65116
$endgroup$
$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21
add a comment |
$begingroup$
Le $ X sim Pois(5) , Y sim Pois(10) $ both independent. Suppose I draw and rectangle with width $ X $ and length $ Y $. Suppose the circumference of the rectangle is $ 28 $ what is $ Var(X) $ ?
First I set another variable $ S $ to be the circumference . So $ S = 2X + 2Y $
$$ Var(X) = E[Var(X|S=28)] + Var(E[X|S = 28]) = \E[Var(X|Y=14-X)] + Var(E[X|Y=14-X]) $$
I'm really having hard time grasping this concept, what is $ E[X|S = 28 ] $ ? and what is the variance of such variable? Maybe I should set new variable $ T = (X|S=28) $ and try to understand what distribution it has?
conditional-expectation poisson-distribution variance
asked Jan 18 at 13:14
bm1125bm1125
65116
$endgroup$
Le $ X sim Pois(5) , Y sim Pois(10) $ both independent. Suppose I draw and rectangle with width $ X $ and length $ Y $. Suppose the circumference of the rectangle is $ 28 $ what is $ Var(X) $ ?
First I set another variable $ S $ to be the circumference . So $ S = 2X + 2Y $
$$ Var(X) = E[Var(X|S=28)] + Var(E[X|S = 28]) = \E[Var(X|Y=14-X)] + Var(E[X|Y=14-X]) $$
I'm really having hard time grasping this concept, what is $ E[X|S = 28 ] $ ? and what is the variance of such variable? Maybe I should set new variable $ T = (X|S=28) $ and try to understand what distribution it has?
conditional-expectation poisson-distribution variance
conditional-expectation poisson-distribution variance
asked Jan 18 at 13:14
bm1125bm1125
65116
asked Jan 18 at 13:14
bm1125bm1125
65116
asked Jan 18 at 13:14
bm1125bm1125
65116
asked Jan 18 at 13:14
bm1125bm1125
65116
asked Jan 18 at 13:14
bm1125bm1125
65116
65116
$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21
add a comment |
$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21
$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe that for $k=0,1,2,dots$ we have the following conditional probabilities : $$p_k=P(X=kmid S=28)=frac{P(X=kwedge S=28)}{P(S=28)}=frac{P(X=kwedge X+Y=14)}{P(X+Y=14)}$$
Here $sum_kp_k=1$ so we can speak of a distribution.
In that context there is a variance which can be written as:$$sum_kp_kk^2-left(sum_kp_kkright)^2$$This on base of the general identity $mathsf{Var}(Z)=mathbb EZ^2-(mathbb EZ)^2$.
This is actually the variance that you are after and can be denoted as $mathsf{Var}(Xmid S=28)$.
Further here $sum_kp_kk$ is the expectation and can be denoted as $mathbb E[Xmid S=28]$. So it is not a random variable but a real number, so that your question "what is the variance of such variable" can only be answered with: its variance is $0$.
answered Jan 18 at 13:38
drhabdrhab
102k545136
$endgroup$
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Observe that for $k=0,1,2,dots$ we have the following conditional probabilities : $$p_k=P(X=kmid S=28)=frac{P(X=kwedge S=28)}{P(S=28)}=frac{P(X=kwedge X+Y=14)}{P(X+Y=14)}$$
Here $sum_kp_k=1$ so we can speak of a distribution.
In that context there is a variance which can be written as:$$sum_kp_kk^2-left(sum_kp_kkright)^2$$This on base of the general identity $mathsf{Var}(Z)=mathbb EZ^2-(mathbb EZ)^2$.
This is actually the variance that you are after and can be denoted as $mathsf{Var}(Xmid S=28)$.
Further here $sum_kp_kk$ is the expectation and can be denoted as $mathbb E[Xmid S=28]$. So it is not a random variable but a real number, so that your question "what is the variance of such variable" can only be answered with: its variance is $0$.
answered Jan 18 at 13:38
drhabdrhab
102k545136
$endgroup$
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
add a comment |
$begingroup$
Observe that for $k=0,1,2,dots$ we have the following conditional probabilities : $$p_k=P(X=kmid S=28)=frac{P(X=kwedge S=28)}{P(S=28)}=frac{P(X=kwedge X+Y=14)}{P(X+Y=14)}$$
Here $sum_kp_k=1$ so we can speak of a distribution.
In that context there is a variance which can be written as:$$sum_kp_kk^2-left(sum_kp_kkright)^2$$This on base of the general identity $mathsf{Var}(Z)=mathbb EZ^2-(mathbb EZ)^2$.
This is actually the variance that you are after and can be denoted as $mathsf{Var}(Xmid S=28)$.
Further here $sum_kp_kk$ is the expectation and can be denoted as $mathbb E[Xmid S=28]$. So it is not a random variable but a real number, so that your question "what is the variance of such variable" can only be answered with: its variance is $0$.
answered Jan 18 at 13:38
drhabdrhab
102k545136
$endgroup$
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
add a comment |
$begingroup$
Observe that for $k=0,1,2,dots$ we have the following conditional probabilities : $$p_k=P(X=kmid S=28)=frac{P(X=kwedge S=28)}{P(S=28)}=frac{P(X=kwedge X+Y=14)}{P(X+Y=14)}$$
Here $sum_kp_k=1$ so we can speak of a distribution.
In that context there is a variance which can be written as:$$sum_kp_kk^2-left(sum_kp_kkright)^2$$This on base of the general identity $mathsf{Var}(Z)=mathbb EZ^2-(mathbb EZ)^2$.
This is actually the variance that you are after and can be denoted as $mathsf{Var}(Xmid S=28)$.
Further here $sum_kp_kk$ is the expectation and can be denoted as $mathbb E[Xmid S=28]$. So it is not a random variable but a real number, so that your question "what is the variance of such variable" can only be answered with: its variance is $0$.
answered Jan 18 at 13:38
drhabdrhab
102k545136
$endgroup$
Observe that for $k=0,1,2,dots$ we have the following conditional probabilities : $$p_k=P(X=kmid S=28)=frac{P(X=kwedge S=28)}{P(S=28)}=frac{P(X=kwedge X+Y=14)}{P(X+Y=14)}$$
Here $sum_kp_k=1$ so we can speak of a distribution.
In that context there is a variance which can be written as:$$sum_kp_kk^2-left(sum_kp_kkright)^2$$This on base of the general identity $mathsf{Var}(Z)=mathbb EZ^2-(mathbb EZ)^2$.
This is actually the variance that you are after and can be denoted as $mathsf{Var}(Xmid S=28)$.
Further here $sum_kp_kk$ is the expectation and can be denoted as $mathbb E[Xmid S=28]$. So it is not a random variable but a real number, so that your question "what is the variance of such variable" can only be answered with: its variance is $0$.
answered Jan 18 at 13:38
drhabdrhab
102k545136
answered Jan 18 at 13:38
drhabdrhab
102k545136
answered Jan 18 at 13:38
drhabdrhab
102k545136
answered Jan 18 at 13:38
drhabdrhab
102k545136
102k545136
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
add a comment |
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
$begingroup$
Why $ sum_k pk = 1 $ it is not clear to me?
$endgroup$
– bm1125
Jan 18 at 14:08
1
1
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
Because $sum_kP (X=kwedge S=28)=P (S=28) $.
$endgroup$
– drhab
Jan 18 at 14:35
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
And why is the variance is $ 0 $ ? that doesn't make any sense. I mean if $ X + Y =14 $ each of the variable may get any value between $ 0 $ to $ 14 $ ...
$endgroup$
– bm1125
Jan 18 at 14:37
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
$begingroup$
Not the variance that you are asked to find is $0$. However you wondered "what is the variance of $E[Xmid S=28]$?" That one is $0$ because $E[Xmid S=28]$ is a real number, so is at most a degenerated random variable.
$endgroup$
– drhab
Jan 18 at 15:18
add a comment |
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$begingroup$
You are looking for $text{Var}(X)$ or $text{Var}(X|X+Y=14)?$ They are totally different. Also the law of tatal variance should be $$text{Var}(X)=E[text{Var}(X|S)]+text{Var}(E[X|S]).$$ (It is different from yours.) You can't calculate it from $E[X|S=28]$ and $text{Var}(X|S=28)$ since they are constants. And you cannot also get $text{Var}(X|S=28)$ and $E[X|S=28]$ from $E[text{Var}(X|S)]$ or $text{Var}(E[X|S])$.
$endgroup$
– Song
Jan 18 at 13:55
$begingroup$
I'm looking for $ Var(X|X+Y=14) $
$endgroup$
– bm1125
Jan 18 at 14:21