Suppose that $X$ is a random variable with $E(X^{n})=3^{n}(n+1)!$ for $n \geq 1$. Find the distribution of $X$.

Question

I tried using the moment generating function. I got: $M_{X}(t)=E(e^{tX})=\Sigma_{n=0}^{\infty} (n+1)(3t)^{n}$. But I don't understand which distribution would it be.

Answer 1

As you've found

$$M_X(t) = \sum_{n = 0}^{\infty} (n+1)(3t)^n$$

We observe that this is almost the same power series as

$$\frac{1}{(1-x)^2} = \sum_{n = 0}^\infty (n+1)x^n$$

Thus $$M_X(t) = \frac{1}{(1-3t)^2}$$

I'll assume that the probability density function $f_X(x)$ we are looking for is defined to be $0$ for negative values of $x$.

If we expand the definition of the moment-generating function

$$\begin{align} M_X(t) = E(e^{tX}) = \int_{-\infty}^\infty e^{tx}f_X(x)dx &= \int_{-\infty}^0e^{tx} \cdot 0dx + \int_0^\infty e^{tx}f_X(x)dx \\ &= \int_0^\infty e^{tx}f_X(x)dx \end{align}$$

We can observe that $M_X(t) = \mathcal{L}\{f_X(x)\}(-t)$, where $\mathcal{L}\{f_X(x)\}(-t)$ is the one-sided Laplace transform of $f_X(x)$ evaluated at $-t$.

$$\begin{align} M_X(t) &= \mathcal{L}\{f_X(x)\}(-t) = \frac{1}{(1-3t)^2} \\ &\implies \mathcal{L}\{f_X(x)\}(t) = \frac{1}{(1+3t)^2} \\ &\implies f_X(x) = \mathcal{L}^{-1} \bigg\{\frac{1}{(1+3t)^2}\bigg\} \end{align}$$

Which we can easily compute

$$\begin{align} \mathcal{L}^{-1} \bigg\{ \frac{1}{(1+3t)^2} \bigg\} &= \frac{1}{9} \mathcal{L}^{-1} \bigg\{ \frac{1}{(t + \frac{1}{3})^2} \bigg\} \\ &= \frac{1}{9} e^{-x/3} \mathcal{L}^{-1} \bigg\{ \frac{1}{t^2} \bigg\} \\ &= \frac{1}{9} e^{-x/3} x \end{align}$$

Thus the probability density function $f_X(x)$ associated with the moment-generating function $M_X(t)$ assuming only positive values for $x$ is

$$f_X(x) = \frac{1}{9} x e^{-x/3}$$

Which turns out to be a Gamma distribution with $\theta = 3$ and $k = 2$ as @Amir mentioned.