$Var[W]$ when $W = X - 2Y$
$begingroup$
I am trying to solve $Var[W]$ when $W = X - 2Y$ for independent discrete RVs $X$ & $Y$:
$$E[X] = 9, Var[X] = 4; E[Y] = 2, Var[Y] = 1.$$
I understand
$$ Var[W] = Var[X] + Var[2Y]$$
$$= Var[X] + 4Var[Y]$$
$$ = 4 + 4*1 = 8$$
But I am trying to answer the same using:
$$ Var[W] = E[W^2] - E[W]^2$$
I think
$$E[W] = 1 implies E[W]^2 = 1$$
But when I calculate
$$ E[(X-2Y)^2] = E[X^2 - 4XY + 4Y^2]$$
$$ = E[X^2] - 4E[X]E[Y] + 4[Y^2]$$
$$ = 85 -72 + 20 = 33$$
$$ implies Var[W] = 32$$
Which I don't think is correct.
Where have I gone wrong the second calculation?
Thanks
probability statistics random-variables variance expected-value
asked Dec 13 '18 at 23:24
number8number8
9318
$endgroup$
add a comment |
$begingroup$
I am trying to solve $Var[W]$ when $W = X - 2Y$ for independent discrete RVs $X$ & $Y$:
$$E[X] = 9, Var[X] = 4; E[Y] = 2, Var[Y] = 1.$$
I understand
$$ Var[W] = Var[X] + Var[2Y]$$
$$= Var[X] + 4Var[Y]$$
$$ = 4 + 4*1 = 8$$
But I am trying to answer the same using:
$$ Var[W] = E[W^2] - E[W]^2$$
I think
$$E[W] = 1 implies E[W]^2 = 1$$
But when I calculate
$$ E[(X-2Y)^2] = E[X^2 - 4XY + 4Y^2]$$
$$ = E[X^2] - 4E[X]E[Y] + 4[Y^2]$$
$$ = 85 -72 + 20 = 33$$
$$ implies Var[W] = 32$$
Which I don't think is correct.
Where have I gone wrong the second calculation?
Thanks
probability statistics random-variables variance expected-value
asked Dec 13 '18 at 23:24
number8number8
9318
$endgroup$
add a comment |
$begingroup$
I am trying to solve $Var[W]$ when $W = X - 2Y$ for independent discrete RVs $X$ & $Y$:
$$E[X] = 9, Var[X] = 4; E[Y] = 2, Var[Y] = 1.$$
I understand
$$ Var[W] = Var[X] + Var[2Y]$$
$$= Var[X] + 4Var[Y]$$
$$ = 4 + 4*1 = 8$$
But I am trying to answer the same using:
$$ Var[W] = E[W^2] - E[W]^2$$
I think
$$E[W] = 1 implies E[W]^2 = 1$$
But when I calculate
$$ E[(X-2Y)^2] = E[X^2 - 4XY + 4Y^2]$$
$$ = E[X^2] - 4E[X]E[Y] + 4[Y^2]$$
$$ = 85 -72 + 20 = 33$$
$$ implies Var[W] = 32$$
Which I don't think is correct.
Where have I gone wrong the second calculation?
Thanks
probability statistics random-variables variance expected-value
asked Dec 13 '18 at 23:24
number8number8
9318
$endgroup$
I am trying to solve $Var[W]$ when $W = X - 2Y$ for independent discrete RVs $X$ & $Y$:
$$E[X] = 9, Var[X] = 4; E[Y] = 2, Var[Y] = 1.$$
I understand
$$ Var[W] = Var[X] + Var[2Y]$$
$$= Var[X] + 4Var[Y]$$
$$ = 4 + 4*1 = 8$$
But I am trying to answer the same using:
$$ Var[W] = E[W^2] - E[W]^2$$
I think
$$E[W] = 1 implies E[W]^2 = 1$$
But when I calculate
$$ E[(X-2Y)^2] = E[X^2 - 4XY + 4Y^2]$$
$$ = E[X^2] - 4E[X]E[Y] + 4[Y^2]$$
$$ = 85 -72 + 20 = 33$$
$$ implies Var[W] = 32$$
Which I don't think is correct.
Where have I gone wrong the second calculation?
Thanks
probability statistics random-variables variance expected-value
probability statistics random-variables variance expected-value
asked Dec 13 '18 at 23:24
number8number8
9318
asked Dec 13 '18 at 23:24
number8number8
9318
edited Dec 13 '18 at 23:30
user614671
asked Dec 13 '18 at 23:24
number8number8
9318
asked Dec 13 '18 at 23:24
number8number8
9318
asked Dec 13 '18 at 23:24
number8number8
9318
9318
add a comment |
add a comment |
1 Answer
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$begingroup$
Your mistake is $E[W]=1$. Note that
$$E[W]=E[X-2Y]=E[X]-2E[Y]=9-2cdot 2=5.$$
The rest is correct except
$$Var(W)=33-5^2=8.$$
Alternatively, use
$$Var(X-2Y)=Var(X)+Var(-2Y)+2Cor(X,-2Y).$$
Since $X$ and $Y$ are independent, $Cor(X,-2Y)=-2Cor(X,Y)=0$. Hence,
$$Var(X-2Y)=Var(X)+Var(-2Y)=Var(X)+4Var(Y)=4+4cdot 1=8.$$
$endgroup$
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your mistake is $E[W]=1$. Note that
$$E[W]=E[X-2Y]=E[X]-2E[Y]=9-2cdot 2=5.$$
The rest is correct except
$$Var(W)=33-5^2=8.$$
Alternatively, use
$$Var(X-2Y)=Var(X)+Var(-2Y)+2Cor(X,-2Y).$$
Since $X$ and $Y$ are independent, $Cor(X,-2Y)=-2Cor(X,Y)=0$. Hence,
$$Var(X-2Y)=Var(X)+Var(-2Y)=Var(X)+4Var(Y)=4+4cdot 1=8.$$
$endgroup$
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
add a comment |
$begingroup$
Your mistake is $E[W]=1$. Note that
$$E[W]=E[X-2Y]=E[X]-2E[Y]=9-2cdot 2=5.$$
The rest is correct except
$$Var(W)=33-5^2=8.$$
Alternatively, use
$$Var(X-2Y)=Var(X)+Var(-2Y)+2Cor(X,-2Y).$$
Since $X$ and $Y$ are independent, $Cor(X,-2Y)=-2Cor(X,Y)=0$. Hence,
$$Var(X-2Y)=Var(X)+Var(-2Y)=Var(X)+4Var(Y)=4+4cdot 1=8.$$
$endgroup$
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
add a comment |
$begingroup$
Your mistake is $E[W]=1$. Note that
$$E[W]=E[X-2Y]=E[X]-2E[Y]=9-2cdot 2=5.$$
The rest is correct except
$$Var(W)=33-5^2=8.$$
Alternatively, use
$$Var(X-2Y)=Var(X)+Var(-2Y)+2Cor(X,-2Y).$$
Since $X$ and $Y$ are independent, $Cor(X,-2Y)=-2Cor(X,Y)=0$. Hence,
$$Var(X-2Y)=Var(X)+Var(-2Y)=Var(X)+4Var(Y)=4+4cdot 1=8.$$
$endgroup$
Your mistake is $E[W]=1$. Note that
$$E[W]=E[X-2Y]=E[X]-2E[Y]=9-2cdot 2=5.$$
The rest is correct except
$$Var(W)=33-5^2=8.$$
Alternatively, use
$$Var(X-2Y)=Var(X)+Var(-2Y)+2Cor(X,-2Y).$$
Since $X$ and $Y$ are independent, $Cor(X,-2Y)=-2Cor(X,Y)=0$. Hence,
$$Var(X-2Y)=Var(X)+Var(-2Y)=Var(X)+4Var(Y)=4+4cdot 1=8.$$
answered Dec 13 '18 at 23:30
user614671
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
add a comment |
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
$begingroup$
Ah yes, sorry I was multiplying $2^2 * 2$ for $2E[Y]$ for some reason. Late at night I guess. Thanks.
$endgroup$
– number8
Dec 13 '18 at 23:33
add a comment |
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