How to prove that the geometric distribution converges to the exponential distribution?

Question

We have given $X_n$ to be a sequence of random variables s.t. $X_n$ has the geometric distribution with parameter $\frac{1}{n}$. I need to show that the distribution of $P_{X_n}$ converges to an exponential distribution. And I also need to find the parameter of this exponential distribution. As a hint they gave us that we may us the formula for geometric series and also the definition of $e^x$ in terms of a limit.

I know that I need to show that $X_n$ converges to $X$ in distribution. By definition we need to show that for all continuous and bounded function $f:\Bbb{R}\rightarrow \Bbb{R}$ we have $$E(f(X_n))\rightarrow E(f(X))$$ so we only had this statement/definition until now therefore we need to work with it.

My idea was to using the definition of the integral by the Riemann sum. So I know that I should get something like $$\int_0^\infty f(x)\lambda e^{-\lambda }dx=\lim_{n\rightarrow \infty} \sum_{j=0}^\infty \frac{1}{n}f\left(\frac{j}{n}\right) \lambda e^{-\lambda \frac{j}{n}}$$ But now I wanted to take $\lambda =1$ because at the end I should get $$\sum_{k=1}^\infty f(k)\frac{1}{n} \left(1-\frac{1}{n}\right)^{k-1}$$But I somehow don't see how to proceed. Could maybe someone help me?

Didn't you forget the normalization? It would make more sense to show that $\color{red}{\frac{\color{black}{X_n}}n}$ tends to the exponential distribution. — nejimban, Apr 23 '22 at 19:50
You wrote “I need to show that the distribution of $P_{X_n}$ converges to an exponential distribution.” You have to consider $P_{\frac{X_n}n}$ instead. So $f(k)$ in the sum there should be $f(\frac kn)$… — nejimban, Apr 23 '22 at 19:53
Note that $\Bbb E[X_n]=n$. So it makes (much) more sense to look at $\frac{X_n}n$ to expect something in the limit $n\to\infty$. — nejimban, Apr 23 '22 at 19:56
@nejimban okey thank you. But then I have $E(f(\frac{X_n}{n}))=\sum_{k=1}^\infty f(\frac{k}{n})\frac{1}{n}(1-\frac{1}{n})^{k-1}$ right? — user1294729, Apr 23 '22 at 20:00
@nejimban but then I Stoll do not get how to get from $(1-\frac{1}{n})^{k-1}$ to $e^{-x}$ — user1294729, Apr 23 '22 at 20:01
Does this answer your question? Geometric distributions converging to exponential distribution — nejimban, Apr 23 '22 at 20:10

score 1 · Accepted Answer · answered Apr 23 '22 at 20:04

1

Including the (necessary) normalization, we simply transform the sum into an integral: $$\Bbb E\!\left[f\!\left(\frac{X_n}n\right)\right]=\sum_{k=1}^\infty\frac1n\left(1-\frac1n\right)^{k-1}f\!\left(\frac kn\right) =\int_0^\infty\frac1n\left(1-\frac1n\right)^{\lfloor x\rfloor}f\!\left(\frac{\lfloor x\rfloor+1}n\right)\,\mathrm dx,$$ where $\lfloor\cdot\rfloor$ denotes integer part. The change of variable $x=ny$ gives $$\Bbb E[f(X_n)]=\int_0^\infty\left(1-\frac1n\right)^{\!\lfloor ny\rfloor} f\!\left(\frac{\lfloor ny\rfloor+1}n\right)\mathrm dy.$$ The integrand tends to $\mathrm e^{-y}f(y)$ and is dominated by the integrable function $y\mapsto\mathrm e^{-y}\|f\|_\infty$ on $(0,\infty)$.

answered Apr 23 '22 at 20:04

nejimban

3,663

@nejmban how do you pass from the sum to the integral? – user1294729 Apr 23 '22 at 20:07
$\int_0^\infty g(\lfloor x\rfloor),\mathrm dx=\sum_{k=0}^\infty\int_k^{k+1}g(\lfloor x\rfloor),\mathrm dx$ and $\lfloor x\rfloor=k$ on $[k,k+1)$. – nejimban Apr 23 '22 at 20:09
But do I really need the integral part thing. Because we have never seen this before and so I think it should be solvable without this? – user1294729 Apr 23 '22 at 20:10
Yes, see here, or here. – nejimban Apr 23 '22 at 20:11
I saw both of them but I didn't understand them. – user1294729 Apr 23 '22 at 20:12
Then please explain in your orginal post the points you do not understand. – nejimban Apr 23 '22 at 20:13
so I mean in both of them they use convergence of the distribution function to define convergence in distribution right? But we haven't discussed this characterization yet – user1294729 Apr 23 '22 at 20:15
I don't think one can easily use a “Riemann sum” argument in your original approach as the integral has the unbounded domain $(0,\infty)$. Rewriting the sum directly as an integral (as I did) circumvents this issue. – nejimban Apr 23 '22 at 20:24
ah okey but then to take the limit inside you use the monotone convergence theorem? – user1294729 Apr 23 '22 at 20:25
Rather, the dominated convergence theorem, because the test function $f$ need not be monotonic. – nejimban Apr 23 '22 at 20:26
ah sure yes I see thanks! – user1294729 Apr 23 '22 at 20:28

How to prove that the geometric distribution converges to the exponential distribution?

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