If $X$ is a random variable, then it's moment generating function is the function

$$ t \rightarrow E(\exp(tX)) $$

which takes values in $]0,+\infty[$.

For a sequence of random variables $X_n$ point convergence of their MGFs to the MGF of a limit distribution intuitively seems like an extremely strong convergence.

In particular, it should be stronger than the weak convergence: so that the following statement is true:

  • If there exists $r$ such that: $\forall t \in [-r,r]$ then $E(\exp(tX_n)) \rightarrow E(\exp(tX))$, then $X_n \rightarrow X$ weakly

I have found three references that say that this is true, but don't offer detailed proofs (Kallenberg, second edition theorem 5.22, http://www.math.uni-bremen.de/~osius/download/papers/MAP33Osius.pdf lemma 2, http://www.randomservices.org/random/expect/Generating.html). Can anybody give a reference for a full detailed proof ?

Edit: I have finally found a proof in Billingsley. see answer below

  • $\begingroup$ For characteristic functions the Levy (aka Levy-Cramer) continuity theorem shows that $X_n \rightarrow_d X \iff \phi_{X_n}(t) \rightarrow \phi_X(t)$ for all $t \in \mathbb R$. A proof can be found in Jun Shao's Mathematical Statistics, among other places (page 56 in 2nd ed). $\endgroup$
    – jld
    Commented Nov 4, 2015 at 15:05
  • $\begingroup$ The relationship between pointwise convergence of the characteristic function and weak convergence is indeed well-known. This question is about pointwise convergence of the moment-generating function which, as far as I can tell, is different enough. $\endgroup$ Commented Nov 4, 2015 at 16:32

1 Answer 1


I finally found a proof I understood. I took it from: Billingsley "Probability and Measure". In order to be thorough, I reproduce the full argument here.

Thm: If $X_n$ is a sequence of random variables for which:

  • The MGF is defined for $t \in [-r,r]$

  • The MGF converges pointwise for $t \in [-r,r]$ to the MGF of $X$

then $X_n \rightarrow X$ in weak convergence, and further all moments of $X_n$ converge to the corresponding moment of $X$

Proof: First we prove that the sequence $X_n$ is tight. Since $E( \exp(-r X_n) + \exp(r X_n) )$ converges, it's bounded. From this boundedness we can prove tightness of the sequence $X_n$.

Since the sequence is tight, we can extract a subsequence $X_{n_k}$ which converges weakly to some limit $\tilde X$. By continuity, $\tilde X$ has a MGF which is equal to that of $X$. Since the MGF characterizes a random variable $\tilde X = X$ (see lemma below)

Every convergent subsequence converges to $X$ and $X_n$ is tight, so $X_n \rightarrow X$ weakly (Billingsley Thm 25.10, corollary)

Lemma: the MGF characterizes the distribution of a random variable: If $X$ and $Y$ have the same MGF for $t \in [-r,r]$, then $X$ and $Y$ have the same distribution

Proof: If the MGF is defined over $[-r,r]$, then it is analytical over $]-r,r[$. We can then extend it to the complex plane for $Re(z) \in ]-r,r[$ and this extension is unique. Note $\psi(z)$ that extension. $\phi(t)=\psi(it)$ is the characteristic function of $X$ which uniquely determines it.

  • 2
    $\begingroup$ If this answers your question, you can click the green check mark next to it and it will be marked as answered. $\endgroup$
    – Sycorax
    Commented Nov 10, 2015 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.