Answer :
Answer:
[tex]E[X^2]= \frac{2!}{2^1 1!}= 1[/tex]
[tex]Var(X^2)= 3-(1)^2 =2[/tex]
Step-by-step explanation:
For this case we can use the moment generating function for the normal model given by:
[tex] \phi(t) = E[e^{tX}][/tex]
And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:
[tex]\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx[/tex]
And we have that the moment generating function can be write like this:
[tex]\phi(t) = e^{\frac{t^2}{2}[/tex]
And we can write this as an infinite series like this:
[tex]\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...[/tex]
And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:
[tex]E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....][/tex]
[tex]E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...[/tex]
and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:
[tex]\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}[/tex]
And then we have this:
[tex]E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...[/tex]
And then we can find the [tex]E[X^2][/tex]
[tex]E[X^2]= \frac{2!}{2^1 1!}= 1[/tex]
And we can find the variance like this :
[tex]Var(X^2) = E[X^4]-[E(X^2)]^2[/tex]
And first we find:
[tex]E[X^4]= \frac{4!}{2^2 2!}= 3[/tex]
And then the variance is given by:
[tex]Var(X^2)= 3-(1)^2 =2[/tex]
Answer:
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Step-by-step explanation
For this case we can use the moment generating function for the normal model given by:
And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:
And we have that the moment generating function can be write like this:
And we can write this as an infinite series like this:
And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:
and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:
And then we have this:
And then we can find the
And we can find the variance like this :
And first we find:
And then the variance is given by: