Convergence of expectations of a sequence of exponential random variables.

Question

Suppose $\{X_n\}$ is a sequence of exponentially distributed random variables, where $X_n$ has mean $1/\lambda_n$. Suppose that $\lim_{n\to\infty}\lambda_n = \lambda>0$. Let $X$ be exponentially distributed with mean $1/\lambda$. Show that $\lim_{n\to\infty}\mathbb E[X_n^m] = \mathbb E[X^m]$ for every integer $m\geqslant 1$.

This looks like a straightforward question - we have $\mathbb E[X_n^m] = \frac{m!}{\lambda_n^m}$ and $\mathbb E[X^m] = \frac{m!}{\lambda^m}$, and clearly $$\lim_{n\to\infty} \frac{m!}{\lambda_n^m} = \frac{m!}{\lambda^m}. $$

The solution provided for this exercise invokes the Skorohod representation theorem, in order to use dominated convergence. This seems entirely unnecessary to me - is there some subtle detail I am missing here?

Answer 1

This question has been answered in the discussion within the comments:

No. It looks like the solution writer missed the not-so-subtle point that the moments of the exponential distribution can be calculated explicitly. – John Dawkins Jul 23 at 18:26

And/or the fact that each $X_n$ is distributed like $Y/λ_n$ and that $X$ is distributed like $Y/λ$, for some exponential random variable $Y$ of parameter $1$. – Did Jul 23 at 19:34.

Posting this as a community wiki answer so that the question may be removed from the list of unanswered questions.