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Problem statement:

Let $X$ be a discrete random variable. Given that $E(X) = 0$, $E(X^2) = 2$ and $E(X^4) = 4$, find the moment-generating function (MGF) for $X$.

Now I know that the general formula for the MGF is $M_X(t) = \sum_{i = 0}^{\infty} E(X^i)\frac{t^i}{i!}$. However, I am provided with only three moment values, not a general formula. Perhaps, the fact that $X$ is discrete is an important clue.

How do I proceed with this problem?

RedDocMD
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2 Answers2

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Hint : $Var(X^2) = E[(X^2)^2] - (E[X^2])^2 = 0$. Thus, $X^2$ is a constant, but $E[X^2] = 2$ so $X^2 = 2$ almost surely. Then $X$ can take only two values, $\pm \sqrt 2$, each with some probability. Find these using the given constraints, and then find $E[e^{tX}]$, as desired.

Sarvesh Ravichandran Iyer
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  • Why almost surely. Why not surely? – Upstart Jun 16 '22 at 19:22
  • @Upstart Thank you for the question. A random variable is a map from a sample space $\Omega \to \mathbb R$. Now, any statement that holds surely is something that will hold for every sample element (in contrast with something holding only almost surely). Generally, while defining random variables, we tend to ignore the sample space because it's possible to have two random variables with the same distribution, but which are different as functions on the sample space. – Sarvesh Ravichandran Iyer Jun 17 '22 at 04:26
  • We can take an example to clarify this. Let $Y$ be a normal random variable (or any continuous random variable with support on $\mathbb R$). Let $X_1 = -\sqrt 2$ if $Y > 0$ and $+\sqrt{2}$ otherwise. On the other hand, let $X_2 = X_1$ except $X_2=0$ if $Y=0$. Then $X_2^2$ and $X_1^2$ have the same distribution, but will be different at every sample space element for which $Y=0$. To avoid trivialities like this, such equalities can only be respected "almost surely". – Sarvesh Ravichandran Iyer Jun 17 '22 at 04:30
  • Just so you mentioned about random variable being a function such that $X^{-1}(w)\in \sigma-$ algebra. Why not define $X$ from $\sigma-$algebra to $\mathbb{R}$ – Upstart Jun 17 '22 at 07:14
  • @Upstart I'm not sure about what you mean by $X^{-1}(\omega)$ , because usually you take $X^{-1}(B)$ where $B \subset R$ is Borel, and this should belong to the sigma-algebra associated to $\Omega$. Nevertheless, if I think I'm getting what you're saying, not every set in the sigma algebra needs to be of the form $X^{-1}(B)$ for $B$ Borel. Defining $X$ as a function from the sigma-algebra to $\mathbb R$ is forcing every possible event to be determined by $X$, which is not always possible. Having said that , I find the idea interesting, but I don't know what one can do with it. – Sarvesh Ravichandran Iyer Jun 17 '22 at 07:23
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$\newcommand{\E}{\mathbb{E}}$Hint: Let $Z=X^2$. Then we are given that $\E\left[Z^2\right] = \left(\E\left[Z\right]\right)^2$. What can we conclude?

Minus One-Twelfth
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