Mixing
Moment-generating function of $Z:=X_1X_2+X_3X_4$
$begingroup$
Let $X_1,X_2,X_3,X_4$ be four indipendent random variable with normal distribution of mean 0 and variance 1. The exercise asks me to calculate the moment-generating function of $X_1X_2$. I was able to do it and I found that $$M_{X_1X_2}(z) = displaystylefrac{1}{sqrt{1-z^2}}$$
I believe that this result is correct since I found a similar question here. The exercise goes on and ask to prove that the moment-generating function of $Z:=X_1X_2+X_3X_4$ is $$M_Z(t)= displaystylefrac{1}{1+t^2}$$
What I write was:
$$M_Z(t)=Bbb E[e^{tZ}]=Bbb E[e^{tX_1X_2+tX_3X_4}]=Bbb E[e^{tX_1X_2}e^{tX_3X_4}] overset{indip.}{=} Bbb E[e^{tX_1X_2}]Bbb E[e^{tX_3X_4}]=displaystyle left(frac{1}{sqrt{1-t^2}}right)^2=frac{1}{1-t^2}$$
Am I missing something or there is an error in the exercise's text?
proof-verification random-variables moment-generating-functions
asked Jan 27 at 10:54
edo1998edo1998
419113
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,X_3,X_4$ be four indipendent random variable with normal distribution of mean 0 and variance 1. The exercise asks me to calculate the moment-generating function of $X_1X_2$. I was able to do it and I found that $$M_{X_1X_2}(z) = displaystylefrac{1}{sqrt{1-z^2}}$$
I believe that this result is correct since I found a similar question here. The exercise goes on and ask to prove that the moment-generating function of $Z:=X_1X_2+X_3X_4$ is $$M_Z(t)= displaystylefrac{1}{1+t^2}$$
What I write was:
$$M_Z(t)=Bbb E[e^{tZ}]=Bbb E[e^{tX_1X_2+tX_3X_4}]=Bbb E[e^{tX_1X_2}e^{tX_3X_4}] overset{indip.}{=} Bbb E[e^{tX_1X_2}]Bbb E[e^{tX_3X_4}]=displaystyle left(frac{1}{sqrt{1-t^2}}right)^2=frac{1}{1-t^2}$$
Am I missing something or there is an error in the exercise's text?
proof-verification random-variables moment-generating-functions
asked Jan 27 at 10:54
edo1998edo1998
419113
$endgroup$
1
$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
1
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59
add a comment |
$begingroup$
Let $X_1,X_2,X_3,X_4$ be four indipendent random variable with normal distribution of mean 0 and variance 1. The exercise asks me to calculate the moment-generating function of $X_1X_2$. I was able to do it and I found that $$M_{X_1X_2}(z) = displaystylefrac{1}{sqrt{1-z^2}}$$
I believe that this result is correct since I found a similar question here. The exercise goes on and ask to prove that the moment-generating function of $Z:=X_1X_2+X_3X_4$ is $$M_Z(t)= displaystylefrac{1}{1+t^2}$$
What I write was:
$$M_Z(t)=Bbb E[e^{tZ}]=Bbb E[e^{tX_1X_2+tX_3X_4}]=Bbb E[e^{tX_1X_2}e^{tX_3X_4}] overset{indip.}{=} Bbb E[e^{tX_1X_2}]Bbb E[e^{tX_3X_4}]=displaystyle left(frac{1}{sqrt{1-t^2}}right)^2=frac{1}{1-t^2}$$
Am I missing something or there is an error in the exercise's text?
proof-verification random-variables moment-generating-functions
asked Jan 27 at 10:54
edo1998edo1998
419113
$endgroup$
Let $X_1,X_2,X_3,X_4$ be four indipendent random variable with normal distribution of mean 0 and variance 1. The exercise asks me to calculate the moment-generating function of $X_1X_2$. I was able to do it and I found that $$M_{X_1X_2}(z) = displaystylefrac{1}{sqrt{1-z^2}}$$
I believe that this result is correct since I found a similar question here. The exercise goes on and ask to prove that the moment-generating function of $Z:=X_1X_2+X_3X_4$ is $$M_Z(t)= displaystylefrac{1}{1+t^2}$$
What I write was:
$$M_Z(t)=Bbb E[e^{tZ}]=Bbb E[e^{tX_1X_2+tX_3X_4}]=Bbb E[e^{tX_1X_2}e^{tX_3X_4}] overset{indip.}{=} Bbb E[e^{tX_1X_2}]Bbb E[e^{tX_3X_4}]=displaystyle left(frac{1}{sqrt{1-t^2}}right)^2=frac{1}{1-t^2}$$
Am I missing something or there is an error in the exercise's text?
proof-verification random-variables moment-generating-functions
proof-verification random-variables moment-generating-functions
asked Jan 27 at 10:54
edo1998edo1998
419113
asked Jan 27 at 10:54
edo1998edo1998
419113
asked Jan 27 at 10:54
edo1998edo1998
419113
asked Jan 27 at 10:54
edo1998edo1998
419113
asked Jan 27 at 10:54
edo1998edo1998
419113
419113
1
$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
1
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59
add a comment |
1
$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
1
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59
1
1
$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
1
1
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59
add a comment |
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$begingroup$
Result is certainly true : stats.stackexchange.com/q/71126/119261.
$endgroup$
– StubbornAtom
Jan 27 at 11:45
$begingroup$
Thank you @StubbornAtom. I wasn't able to find a reference of this
$endgroup$
– edo1998
Jan 27 at 12:54
1
$begingroup$
@StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=phi(-it)$ (where $phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$).
$endgroup$
– edo1998
Jan 27 at 13:44
$begingroup$
You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here.
$endgroup$
– StubbornAtom
Jan 27 at 13:57
$begingroup$
@StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it!
$endgroup$
– edo1998
Jan 27 at 13:59