Mixing
How do I calculate $E[X^2Y]$ for discrete (and DEPENDENT) random variable given the joint pmf table?
$begingroup$
So far I have found the marginal pmf for X and Y and proved that the random variable are not independent by showing that;
$P(X=i, y=j)neq P(X=i)P(Y=j)$
The next question asks to calculate;
$E[X^2Y]$
To calculate this I'm using the following property;
$Cov(X,Y) =E(XY)-E(X)E(Y)$
So in relation to my question;
$E(X^2Y) = Cov(X^2,Y)+E(X^2)E(Y)$
I have calculated $E(X^2)$ & $E(Y)$
To calculate the $Cov(X^2,Y)$
can I use the following formula for discrete RV;
$Cov(X,Y)=sum sum_{(x,y)in S} (x-mu_x)(y-mu_y)f(x,y) $
where f(x,y) is the joint pmf
and simply replace the expectation of x with the second moment of x?
ie.
$Cov(X^2,Y)=sum sum_{(x,y)in S} (x^2-mu_x^2)(y-mu_y)f(x,y) $
Any help much appreciated!
probability probability-theory statistics probability-distributions
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
$endgroup$
add a comment |
$begingroup$
So far I have found the marginal pmf for X and Y and proved that the random variable are not independent by showing that;
$P(X=i, y=j)neq P(X=i)P(Y=j)$
The next question asks to calculate;
$E[X^2Y]$
To calculate this I'm using the following property;
$Cov(X,Y) =E(XY)-E(X)E(Y)$
So in relation to my question;
$E(X^2Y) = Cov(X^2,Y)+E(X^2)E(Y)$
I have calculated $E(X^2)$ & $E(Y)$
To calculate the $Cov(X^2,Y)$
can I use the following formula for discrete RV;
$Cov(X,Y)=sum sum_{(x,y)in S} (x-mu_x)(y-mu_y)f(x,y) $
where f(x,y) is the joint pmf
and simply replace the expectation of x with the second moment of x?
ie.
$Cov(X^2,Y)=sum sum_{(x,y)in S} (x^2-mu_x^2)(y-mu_y)f(x,y) $
Any help much appreciated!
probability probability-theory statistics probability-distributions
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
$endgroup$
1
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28
add a comment |
$begingroup$
So far I have found the marginal pmf for X and Y and proved that the random variable are not independent by showing that;
$P(X=i, y=j)neq P(X=i)P(Y=j)$
The next question asks to calculate;
$E[X^2Y]$
To calculate this I'm using the following property;
$Cov(X,Y) =E(XY)-E(X)E(Y)$
So in relation to my question;
$E(X^2Y) = Cov(X^2,Y)+E(X^2)E(Y)$
I have calculated $E(X^2)$ & $E(Y)$
To calculate the $Cov(X^2,Y)$
can I use the following formula for discrete RV;
$Cov(X,Y)=sum sum_{(x,y)in S} (x-mu_x)(y-mu_y)f(x,y) $
where f(x,y) is the joint pmf
and simply replace the expectation of x with the second moment of x?
ie.
$Cov(X^2,Y)=sum sum_{(x,y)in S} (x^2-mu_x^2)(y-mu_y)f(x,y) $
Any help much appreciated!
probability probability-theory statistics probability-distributions
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
$endgroup$
So far I have found the marginal pmf for X and Y and proved that the random variable are not independent by showing that;
$P(X=i, y=j)neq P(X=i)P(Y=j)$
The next question asks to calculate;
$E[X^2Y]$
To calculate this I'm using the following property;
$Cov(X,Y) =E(XY)-E(X)E(Y)$
So in relation to my question;
$E(X^2Y) = Cov(X^2,Y)+E(X^2)E(Y)$
I have calculated $E(X^2)$ & $E(Y)$
To calculate the $Cov(X^2,Y)$
can I use the following formula for discrete RV;
$Cov(X,Y)=sum sum_{(x,y)in S} (x-mu_x)(y-mu_y)f(x,y) $
where f(x,y) is the joint pmf
and simply replace the expectation of x with the second moment of x?
ie.
$Cov(X^2,Y)=sum sum_{(x,y)in S} (x^2-mu_x^2)(y-mu_y)f(x,y) $
Any help much appreciated!
probability probability-theory statistics probability-distributions
probability probability-theory statistics probability-distributions
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
edited Oct 14 '17 at 8:41
S.Tee
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
asked Oct 14 '17 at 8:09
S.TeeS.Tee
12
12
1
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28
add a comment |
1
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28
1
1
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use $E(X^2Y)=sum_isum_jP(X=i, Y=j) i^2j$.
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
$endgroup$
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
add a comment |
$begingroup$
You can do that, but then you must be precise:$$mathsf{Cov}(X^2,Y)=mathbb E(X^2-mathbb EX^2)(Y-mathbb EY)=sum_xsum_y(x^2-mu_{X^2})(y-mu_Y)f(x,y)tag1$$
Note that in $(1)$ we have $mu_{X^2}=mathbb EX^2$ and not $mu_X^2=(mathbb EX)^2$.
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Use $E(X^2Y)=sum_isum_jP(X=i, Y=j) i^2j$.
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
$endgroup$
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
add a comment |
$begingroup$
Use $E(X^2Y)=sum_isum_jP(X=i, Y=j) i^2j$.
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
$endgroup$
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
add a comment |
$begingroup$
Use $E(X^2Y)=sum_isum_jP(X=i, Y=j) i^2j$.
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
$endgroup$
Use $E(X^2Y)=sum_isum_jP(X=i, Y=j) i^2j$.
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
answered Oct 14 '17 at 8:19
SrikantSrikant
31537
31537
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
add a comment |
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
$begingroup$
This could turn out to be a good advice, but is actually not an answer to the question.
$endgroup$
– drhab
Oct 14 '17 at 8:39
add a comment |
$begingroup$
You can do that, but then you must be precise:$$mathsf{Cov}(X^2,Y)=mathbb E(X^2-mathbb EX^2)(Y-mathbb EY)=sum_xsum_y(x^2-mu_{X^2})(y-mu_Y)f(x,y)tag1$$
Note that in $(1)$ we have $mu_{X^2}=mathbb EX^2$ and not $mu_X^2=(mathbb EX)^2$.
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You can do that, but then you must be precise:$$mathsf{Cov}(X^2,Y)=mathbb E(X^2-mathbb EX^2)(Y-mathbb EY)=sum_xsum_y(x^2-mu_{X^2})(y-mu_Y)f(x,y)tag1$$
Note that in $(1)$ we have $mu_{X^2}=mathbb EX^2$ and not $mu_X^2=(mathbb EX)^2$.
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You can do that, but then you must be precise:$$mathsf{Cov}(X^2,Y)=mathbb E(X^2-mathbb EX^2)(Y-mathbb EY)=sum_xsum_y(x^2-mu_{X^2})(y-mu_Y)f(x,y)tag1$$
Note that in $(1)$ we have $mu_{X^2}=mathbb EX^2$ and not $mu_X^2=(mathbb EX)^2$.
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
$endgroup$
You can do that, but then you must be precise:$$mathsf{Cov}(X^2,Y)=mathbb E(X^2-mathbb EX^2)(Y-mathbb EY)=sum_xsum_y(x^2-mu_{X^2})(y-mu_Y)f(x,y)tag1$$
Note that in $(1)$ we have $mu_{X^2}=mathbb EX^2$ and not $mu_X^2=(mathbb EX)^2$.
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
edited Oct 14 '17 at 8:30
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
answered Oct 14 '17 at 8:19
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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1
$begingroup$
If you know $P(X=i,Y=j)$, then can you not use the formula $E[X^2Y] = sum_{i,j} i^2 times j times P(X= i,Y=j)$.
$endgroup$
– астон вілла олоф мэллбэрг
Oct 14 '17 at 8:19
$begingroup$
@астонвіллаолофмэллбэрг there are situations where things become easyer if you work with standardized random variables. Then covariances come in sight. That might explain the preference (not sure of that) of the OP.
$endgroup$
– drhab
Oct 14 '17 at 8:27
$begingroup$
In the equality following "So in relation to my question" the minus-sign must be a plus-sign.
$endgroup$
– drhab
Oct 14 '17 at 8:28