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I tried approaching it from the variance perspective, ie $$\operatorname{var}(X)=e^{t^2}(1-e^{t^2})$$ and then I thought that maybe I should take the integral from $-\infty$ to $+\infty$ of the pdf, but that's not really the right approach either.

I think I'm missing some kind of algebraic manipulation here, if someone wouldn't mind pointing me in the right direction...thanks!

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    $\begingroup$ Hint: what is the second derivative of $E[e^{tX}]$ with respect to $t$? $\endgroup$
    – J.G.
    Sep 19, 2022 at 15:39

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Among others, there are two ways to solve this. First, note that for any random variable X whose moment generating function (MGF) exists, $E(X^2) = \frac{d^2}{dt^2} E(e^{tX})|_{t = 0}$. In words, the second moment is the second derivative of the MGF evaluated at 0. Hence, using this approach, and since you know the form of the MGF, we have

$$ E(X^2) = \frac{d^2}{dt^2} E(e^{tX})|_{t = 0} = \frac{d^2}{dt^2}e^{t^2} = \left(2e^{t^2} + 4t^2 e^{t^2}\right)|_{t = 0} = 2. $$ Another way is if you know the MGF of the normal distribution. Indeed for a normal distribution with mean $\mu$ and variance $\sigma^2$, the MGF is $E(e^{tX}) = e^{\mu t + \sigma^2 t^2 /2}$. So you could view your $E(e^{tX})$ as the MGF of a normal distribution with mean $0$ and variance $2$. Thus, $E(X^2) = Var(X) = 2$. Hope this helps!

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  • $\begingroup$ thanks! this was very clear $\endgroup$ Sep 19, 2022 at 17:09

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