Let x be a Standard Gaussian Random Variable. Find the mean and variance of ax+b. [closed]
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I was looking up probability practice problems, and I came across this question. How would I go about solving this?
probability
asked Jan 19 at 18:45
ryan6297ryan6297
122
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closed as off-topic by StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500 Jan 20 at 7:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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I was looking up probability practice problems, and I came across this question. How would I go about solving this?
probability
asked Jan 19 at 18:45
ryan6297ryan6297
122
$endgroup$
closed as off-topic by StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500 Jan 20 at 7:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
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– chickenNinja123
Jan 19 at 18:52
1
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By applying linearity of expectation and bilinearity of covariance.
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– drhab
Jan 19 at 18:55
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You should not pose the question solely in the title.
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– LoveTooNap29
Jan 19 at 18:55
add a comment |
$begingroup$
I was looking up probability practice problems, and I came across this question. How would I go about solving this?
probability
asked Jan 19 at 18:45
ryan6297ryan6297
122
$endgroup$
I was looking up probability practice problems, and I came across this question. How would I go about solving this?
probability
probability
asked Jan 19 at 18:45
ryan6297ryan6297
122
asked Jan 19 at 18:45
ryan6297ryan6297
122
asked Jan 19 at 18:45
ryan6297ryan6297
122
asked Jan 19 at 18:45
ryan6297ryan6297
122
asked Jan 19 at 18:45
ryan6297ryan6297
122
122
closed as off-topic by StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500 Jan 20 at 7:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500 Jan 20 at 7:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Did, Lord Shark the Unknown, RRL, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
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– chickenNinja123
Jan 19 at 18:52
1
$begingroup$
By applying linearity of expectation and bilinearity of covariance.
$endgroup$
– drhab
Jan 19 at 18:55
$begingroup$
You should not pose the question solely in the title.
$endgroup$
– LoveTooNap29
Jan 19 at 18:55
add a comment |
1
$begingroup$
This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
$endgroup$
– chickenNinja123
Jan 19 at 18:52
1
$begingroup$
By applying linearity of expectation and bilinearity of covariance.
$endgroup$
– drhab
Jan 19 at 18:55
$begingroup$
You should not pose the question solely in the title.
$endgroup$
– LoveTooNap29
Jan 19 at 18:55
1
1
$begingroup$
This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
$endgroup$
– chickenNinja123
Jan 19 at 18:52
$begingroup$
This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
$endgroup$
– chickenNinja123
Jan 19 at 18:52
1
1
$begingroup$
By applying linearity of expectation and bilinearity of covariance.
$endgroup$
– drhab
Jan 19 at 18:55
$begingroup$
By applying linearity of expectation and bilinearity of covariance.
$endgroup$
– drhab
Jan 19 at 18:55
$begingroup$
You should not pose the question solely in the title.
$endgroup$
– LoveTooNap29
Jan 19 at 18:55
$begingroup$
You should not pose the question solely in the title.
$endgroup$
– LoveTooNap29
Jan 19 at 18:55
add a comment |
2 Answers
2
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If $a$ and $b$ are constants, we know that,
$$ E[ax+b]=int_{-infty}^{infty}(ax+b) dF(x)=aE[x]+b,$$
irrespective of what the distribution function is. Since $E[x]=0$ for a standard gaussian r.v., the mean is $E[ax+b]=b.$
As for the variance,
$$E[(ax+b-E[x])^2]=int_{-infty}^{infty}(ax+b -b)^2 dF(x)=a^2int_{-infty}^{infty}x^2 dF(x).$$
Again, as for a standard gaussian we have that $int_{-infty}^{infty}x^2 dF(x)=1,$ the variance of $ax+b$ will be $a^2.$
answered Jan 19 at 19:03
PatricioPatricio
3257
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Hint
Define $y=ax+b$ therefore $$E(y)=E(ax+b)\E(y^2)=EBig ((ax+b)^2Big)\sigma^2_y=E(y^2)-Big(E(y)Big)^2$$
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a$ and $b$ are constants, we know that,
$$ E[ax+b]=int_{-infty}^{infty}(ax+b) dF(x)=aE[x]+b,$$
irrespective of what the distribution function is. Since $E[x]=0$ for a standard gaussian r.v., the mean is $E[ax+b]=b.$
As for the variance,
$$E[(ax+b-E[x])^2]=int_{-infty}^{infty}(ax+b -b)^2 dF(x)=a^2int_{-infty}^{infty}x^2 dF(x).$$
Again, as for a standard gaussian we have that $int_{-infty}^{infty}x^2 dF(x)=1,$ the variance of $ax+b$ will be $a^2.$
answered Jan 19 at 19:03
PatricioPatricio
3257
$endgroup$
add a comment |
$begingroup$
If $a$ and $b$ are constants, we know that,
$$ E[ax+b]=int_{-infty}^{infty}(ax+b) dF(x)=aE[x]+b,$$
irrespective of what the distribution function is. Since $E[x]=0$ for a standard gaussian r.v., the mean is $E[ax+b]=b.$
As for the variance,
$$E[(ax+b-E[x])^2]=int_{-infty}^{infty}(ax+b -b)^2 dF(x)=a^2int_{-infty}^{infty}x^2 dF(x).$$
Again, as for a standard gaussian we have that $int_{-infty}^{infty}x^2 dF(x)=1,$ the variance of $ax+b$ will be $a^2.$
answered Jan 19 at 19:03
PatricioPatricio
3257
$endgroup$
add a comment |
$begingroup$
If $a$ and $b$ are constants, we know that,
$$ E[ax+b]=int_{-infty}^{infty}(ax+b) dF(x)=aE[x]+b,$$
irrespective of what the distribution function is. Since $E[x]=0$ for a standard gaussian r.v., the mean is $E[ax+b]=b.$
As for the variance,
$$E[(ax+b-E[x])^2]=int_{-infty}^{infty}(ax+b -b)^2 dF(x)=a^2int_{-infty}^{infty}x^2 dF(x).$$
Again, as for a standard gaussian we have that $int_{-infty}^{infty}x^2 dF(x)=1,$ the variance of $ax+b$ will be $a^2.$
answered Jan 19 at 19:03
PatricioPatricio
3257
$endgroup$
If $a$ and $b$ are constants, we know that,
$$ E[ax+b]=int_{-infty}^{infty}(ax+b) dF(x)=aE[x]+b,$$
irrespective of what the distribution function is. Since $E[x]=0$ for a standard gaussian r.v., the mean is $E[ax+b]=b.$
As for the variance,
$$E[(ax+b-E[x])^2]=int_{-infty}^{infty}(ax+b -b)^2 dF(x)=a^2int_{-infty}^{infty}x^2 dF(x).$$
Again, as for a standard gaussian we have that $int_{-infty}^{infty}x^2 dF(x)=1,$ the variance of $ax+b$ will be $a^2.$
answered Jan 19 at 19:03
PatricioPatricio
3257
answered Jan 19 at 19:03
PatricioPatricio
3257
answered Jan 19 at 19:03
PatricioPatricio
3257
answered Jan 19 at 19:03
PatricioPatricio
3257
3257
add a comment |
add a comment |
$begingroup$
Hint
Define $y=ax+b$ therefore $$E(y)=E(ax+b)\E(y^2)=EBig ((ax+b)^2Big)\sigma^2_y=E(y^2)-Big(E(y)Big)^2$$
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
Define $y=ax+b$ therefore $$E(y)=E(ax+b)\E(y^2)=EBig ((ax+b)^2Big)\sigma^2_y=E(y^2)-Big(E(y)Big)^2$$
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
Define $y=ax+b$ therefore $$E(y)=E(ax+b)\E(y^2)=EBig ((ax+b)^2Big)\sigma^2_y=E(y^2)-Big(E(y)Big)^2$$
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
Define $y=ax+b$ therefore $$E(y)=E(ax+b)\E(y^2)=EBig ((ax+b)^2Big)\sigma^2_y=E(y^2)-Big(E(y)Big)^2$$
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 18:56
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
1
$begingroup$
This one can even be done without the assumption of $x$ being Gaussian. Simply use the linearity of the expectation and the fact that the variance is "homogeneous of order 2".
$endgroup$
– chickenNinja123
Jan 19 at 18:52
1
$begingroup$
By applying linearity of expectation and bilinearity of covariance.
$endgroup$
– drhab
Jan 19 at 18:55
$begingroup$
You should not pose the question solely in the title.
$endgroup$
– LoveTooNap29
Jan 19 at 18:55