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?
Best Answer
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 $(0,\infty)$.