Linear Combination of Independent Random Variables
I am working on a problem and am a bit stuck
The problem: For independent random variables
X ~ N(-1,3)
Y ~ N(0,2)
Z ~ N(4,1)
Consider U:= 2X - 4Y - Z + 5
Find the Expectation, variance, and MGF for U.
What I have done so far:
Fx(t) = Ee$^{tX}$ = exp(-t + $3t^2over2$)
Fy(t) = Ee$^{tY}$ = exp(0t + t$^2$)
Fz(t) = Ee$^{tZ}$ = exp(4t + $t^2over2$)
Then MGF of U:
Fu(t) = Ee$^{t(2X-4Y-Z+5)}$ = E[e$^{(2t)X}$e$^{(-4t)Y}$e$^{(-t)Z}$e$^{(5t)}$] =
e$^{(5t)}$Ee$^{(2t)X}$Ee$^{(-4t)Y}$Ee$^{(-t)Z}$ = e$^{(5t)}$Fx(2t)Fy(-4t)Fz(-1t) =
exp(5t) exp(-(2t) + $3(2t)^2over2$) exp(0(-4t) + (-4t)$^2$) exp(4(-1t) + $(-1t)^2over2$) =
exp(5t -2t +0t -4t + 6t$^2$ +16t$^2$ + $1over2$$t^2$) = exp(-1t + 22.5t$^2$) =
U ~ N(-1, 22.5)
Then for Expectation:
F'u(t) =
But here is where I am stuck. How do we find the Expectation of U (how do we take this derivative) and how do we find the Variance of U? Also, does the first part look correct?
probability variance moment-generating-functions expected-value
asked Nov 20 at 18:09
Ethan
10012
add a comment |
I am working on a problem and am a bit stuck
The problem: For independent random variables
X ~ N(-1,3)
Y ~ N(0,2)
Z ~ N(4,1)
Consider U:= 2X - 4Y - Z + 5
Find the Expectation, variance, and MGF for U.
What I have done so far:
Fx(t) = Ee$^{tX}$ = exp(-t + $3t^2over2$)
Fy(t) = Ee$^{tY}$ = exp(0t + t$^2$)
Fz(t) = Ee$^{tZ}$ = exp(4t + $t^2over2$)
Then MGF of U:
Fu(t) = Ee$^{t(2X-4Y-Z+5)}$ = E[e$^{(2t)X}$e$^{(-4t)Y}$e$^{(-t)Z}$e$^{(5t)}$] =
e$^{(5t)}$Ee$^{(2t)X}$Ee$^{(-4t)Y}$Ee$^{(-t)Z}$ = e$^{(5t)}$Fx(2t)Fy(-4t)Fz(-1t) =
exp(5t) exp(-(2t) + $3(2t)^2over2$) exp(0(-4t) + (-4t)$^2$) exp(4(-1t) + $(-1t)^2over2$) =
exp(5t -2t +0t -4t + 6t$^2$ +16t$^2$ + $1over2$$t^2$) = exp(-1t + 22.5t$^2$) =
U ~ N(-1, 22.5)
Then for Expectation:
F'u(t) =
But here is where I am stuck. How do we find the Expectation of U (how do we take this derivative) and how do we find the Variance of U? Also, does the first part look correct?
probability variance moment-generating-functions expected-value
asked Nov 20 at 18:09
Ethan
10012
add a comment |
I am working on a problem and am a bit stuck
The problem: For independent random variables
X ~ N(-1,3)
Y ~ N(0,2)
Z ~ N(4,1)
Consider U:= 2X - 4Y - Z + 5
Find the Expectation, variance, and MGF for U.
What I have done so far:
Fx(t) = Ee$^{tX}$ = exp(-t + $3t^2over2$)
Fy(t) = Ee$^{tY}$ = exp(0t + t$^2$)
Fz(t) = Ee$^{tZ}$ = exp(4t + $t^2over2$)
Then MGF of U:
Fu(t) = Ee$^{t(2X-4Y-Z+5)}$ = E[e$^{(2t)X}$e$^{(-4t)Y}$e$^{(-t)Z}$e$^{(5t)}$] =
e$^{(5t)}$Ee$^{(2t)X}$Ee$^{(-4t)Y}$Ee$^{(-t)Z}$ = e$^{(5t)}$Fx(2t)Fy(-4t)Fz(-1t) =
exp(5t) exp(-(2t) + $3(2t)^2over2$) exp(0(-4t) + (-4t)$^2$) exp(4(-1t) + $(-1t)^2over2$) =
exp(5t -2t +0t -4t + 6t$^2$ +16t$^2$ + $1over2$$t^2$) = exp(-1t + 22.5t$^2$) =
U ~ N(-1, 22.5)
Then for Expectation:
F'u(t) =
But here is where I am stuck. How do we find the Expectation of U (how do we take this derivative) and how do we find the Variance of U? Also, does the first part look correct?
probability variance moment-generating-functions expected-value
asked Nov 20 at 18:09
Ethan
10012
I am working on a problem and am a bit stuck
The problem: For independent random variables
X ~ N(-1,3)
Y ~ N(0,2)
Z ~ N(4,1)
Consider U:= 2X - 4Y - Z + 5
Find the Expectation, variance, and MGF for U.
What I have done so far:
Fx(t) = Ee$^{tX}$ = exp(-t + $3t^2over2$)
Fy(t) = Ee$^{tY}$ = exp(0t + t$^2$)
Fz(t) = Ee$^{tZ}$ = exp(4t + $t^2over2$)
Then MGF of U:
Fu(t) = Ee$^{t(2X-4Y-Z+5)}$ = E[e$^{(2t)X}$e$^{(-4t)Y}$e$^{(-t)Z}$e$^{(5t)}$] =
e$^{(5t)}$Ee$^{(2t)X}$Ee$^{(-4t)Y}$Ee$^{(-t)Z}$ = e$^{(5t)}$Fx(2t)Fy(-4t)Fz(-1t) =
exp(5t) exp(-(2t) + $3(2t)^2over2$) exp(0(-4t) + (-4t)$^2$) exp(4(-1t) + $(-1t)^2over2$) =
exp(5t -2t +0t -4t + 6t$^2$ +16t$^2$ + $1over2$$t^2$) = exp(-1t + 22.5t$^2$) =
U ~ N(-1, 22.5)
Then for Expectation:
F'u(t) =
But here is where I am stuck. How do we find the Expectation of U (how do we take this derivative) and how do we find the Variance of U? Also, does the first part look correct?
probability variance moment-generating-functions expected-value
probability variance moment-generating-functions expected-value
asked Nov 20 at 18:09
Ethan
10012
asked Nov 20 at 18:09
Ethan
10012
asked Nov 20 at 18:09
Ethan
10012
asked Nov 20 at 18:09
Ethan
10012
asked Nov 20 at 18:09
Ethan
10012
10012
add a comment |
add a comment |
1 Answer
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Expecation is linear.
$$E[U]= 2E[X] - 4E[Y] - E[Z] + 5$$
Also, since they are independent.
$$Var[U]= 2^2Var[X] + 4^2Var[Y] + Var[Z] $$
Remark:
- Usually we use $F$ for CDF and $M$ for mgf.
$U$ is again a normal distribution and after you find out the mean and variance, figuring out the mgf should be easy.
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
Expecation is linear.
$$E[U]= 2E[X] - 4E[Y] - E[Z] + 5$$
Also, since they are independent.
$$Var[U]= 2^2Var[X] + 4^2Var[Y] + Var[Z] $$
Remark:
- Usually we use $F$ for CDF and $M$ for mgf.
$U$ is again a normal distribution and after you find out the mean and variance, figuring out the mgf should be easy.
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
add a comment |
Expecation is linear.
$$E[U]= 2E[X] - 4E[Y] - E[Z] + 5$$
Also, since they are independent.
$$Var[U]= 2^2Var[X] + 4^2Var[Y] + Var[Z] $$
Remark:
- Usually we use $F$ for CDF and $M$ for mgf.
$U$ is again a normal distribution and after you find out the mean and variance, figuring out the mgf should be easy.
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
add a comment |
Expecation is linear.
$$E[U]= 2E[X] - 4E[Y] - E[Z] + 5$$
Also, since they are independent.
$$Var[U]= 2^2Var[X] + 4^2Var[Y] + Var[Z] $$
Remark:
- Usually we use $F$ for CDF and $M$ for mgf.
$U$ is again a normal distribution and after you find out the mean and variance, figuring out the mgf should be easy.
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
Expecation is linear.
$$E[U]= 2E[X] - 4E[Y] - E[Z] + 5$$
Also, since they are independent.
$$Var[U]= 2^2Var[X] + 4^2Var[Y] + Var[Z] $$
Remark:
- Usually we use $F$ for CDF and $M$ for mgf.
$U$ is again a normal distribution and after you find out the mean and variance, figuring out the mgf should be easy.
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
edited Nov 20 at 18:17
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
answered Nov 20 at 18:12
Siong Thye Goh
99.1k1464117
99.1k1464117
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
add a comment |
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
To clarify though, to find these expectations we need to take the derivative of the MFGs correct? ie. M'x(t) = M'x(0) = EX and M''x(t) = M''x(0) = E(X$^2$)
– Ethan
Nov 20 at 18:25
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
that's just one of the way. but here we are given that $X sim N(-1,3)$, clearly we know $E[X]=-1$ and $Var(X)=3$ and we can compute the second moment from there.
– Siong Thye Goh
Nov 20 at 18:27
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
Ok, that clears things up. Thank you.
– Ethan
Nov 20 at 18:28
add a comment |
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