Define $\{p_k: k \ge 0 \}$ as a probability mass function on a discrete random variable X taking values on $\{0,1,...\}$ then define the generating function $P(x) = E(x^X) = \sum_{i=0}^\infty p_ix^i$ where $x \in [0,1]$. Show that $P'(x) = \sum_{i=0}^\infty p_iix^{i-1}$.

My thoughts:

The dominating convergence theorem says that if $X_n \to X$ and $\exists Y\in L^1$ such that $|X_n| \le Y $, then $E(X_n) \to E(X)$. From the hint given below, lets first define $P_n(x) = \sum_{i=0}^np_ix^i$ (this is clearly a polynomial) then $P_n'(x) = \sum_{i=1}^np_iix^{i-1}$. So we want to show as $n \rightarrow \infty$, $P_n' \rightarrow P'$ uniformly over the compact interval $[0,1]$. Then we want to show $P_n \rightarrow P$ pointwise. I still don't see which part would require the use of dominated convergence theorem in this case since we would be dealing with the convergence of expectations.

Maybe we need to define our sequences here differently such that expectation gives us the result. For example if I let $f_n = x^n$ then $f_n \rightarrow 0$ as $n \rightarrow \infty$ and since $|f_n| \le 1$ by dominate convergence theorem $E(f_n) = P_n(x)\rightarrow 0$, but this doesn't help.


First, observe that it is sufficient to prove the following:

if $f_n$ is a sequence of $C^1$ functions whose derivatives $f_n'$ uniformly converge to a function $g$ on a compact interval, and if $f_n$ converges pointwise to a function $f$,

then $f$ is $C^1$ with derivative $g$.

This is proved by FTC.

Pick a point $x_0$ in your compact interval and write

$f_n (x) = f_n (x_0 )+ \int_{x_0}^{x} f_n'(t)dt$.

We want to take the limit $n \rightarrow \infty$ of both sides.

This is where we need the dominated convergence theorem.

Consider the interval $[0,1-\delta]$ and prove that the function

$\sum_{i=1}^{n} p_i i x^{i-1}$ is bounded by some function $Y(x)$ for all $n$.

  • $\begingroup$ I think I kind of see what your trying to say. But since we are dealing with a discrete random variable in my case, we cannot work with integrals. And the dominated convergences theorem in this case is in terms of expectations. $\endgroup$ – user77404 Nov 10 '13 at 4:41
  • $\begingroup$ I made a edit of my understanding after seeing your insight, but a still a little confused of how expectations come into place. $\endgroup$ – user77404 Nov 10 '13 at 5:04
  • $\begingroup$ We are indeed not working with discrete random variables, but rather with continuous functions (which are polynomials). $\endgroup$ – a12345 Nov 11 '13 at 5:07
  • $\begingroup$ In the question we are given a probability mass function, which is used for discrete RV, but I think I see your point, could you maybe edit your answer to show how we can use expectations in this case, could you see my edit it to my question if the approach I have in mind makes sense? $\endgroup$ – user77404 Nov 11 '13 at 5:29

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