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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.math.uah.edu/stat/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

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  • $\begingroup$ 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 Nov 4 '15 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. $\endgroup$ – Guillaume Dehaene Nov 4 '15 at 16:32
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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.

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    $\begingroup$ If this answers your question, you can click the green check mark next to it and it will be marked as answered. $\endgroup$ – Sycorax says Reinstate Monica Nov 10 '15 at 14:21

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