Let $X_n \sim Beta(k,n+1-k)$, with $k = [\nu n]$ for some $\nu \in (0,1)$.
I want to show that
$$X_n \to \nu $$
almost surely as $n \to \infty$.
My attempt:
I have successfully shown convergence in probability by writing out the probability $P(|X_n – k/n| > \epsilon)$ and applying Markov's inequality. However here I am stuck, and haven't been able to progress further.
I have heard that a beta random variable can be expressed as a ratio of sums of standard exponential distributions:
$$X = \frac{\sum_{i=1}^k Z_i}{\sum_{i=1}^{n+1} Z_i}$$
where $Z_i \sim \exp (1)$.
If this is true (and I think it is, using the relation between Gamma and exponential, and gamma and beta), the Strong Law of Large Numbers could be used to show that $X_n = Y_n/Z_n$ where $Y_n \to k$ almost surely and $Z_n$ converges to $n$ almost surely. But, again, I'm stuck. Since Slutsky's thereom can only be used to show convergence in probability here, not almost sure.
Edit:
According to Wikipedia, if $X_n, Y_n$ converge almost surely to $X,Y$ respectively, then $X_nY_n$ converges to $XY$ almost surely. Hence, since we can show, under the exponential representation, we have $X_n = Y_n / Z_n$ where $Y_n \to k$ and $Z_n \to n$, then applying the continuous mapping theorem to $Z_n$ I believe we can write
$$X_n = Y_n Z^*_n$$
Where $Y_n \to k$ and $Z_n^* \to n^{-1}$. It then follows by the result I found on Wikipedia that $X_n \to k/n$ almost surely. Can someone confirm this proof is correct?
Best Answer
You are really close. I denote $k$ as $k_n:=[n\nu]$ where $[\cdot] $ as you said is the integer part. We can indeed write $$X_n=\frac{\sum_{i=1}^{k_n}Z_i} {\sum_{i=1}^{n+1}Z_i}, $$ where $Z_1,Z_2,...$ are i.i.d. $\exp(1)$ distributed random variables. We know from the SLLN that $$\frac{1}{k_n}{\sum_{i=1}^{k_n}Z_i}\to 1 \ \ \ \text{ a.s. for } n\to\infty, $$ and similarly $$\frac{1}{n+1}{\sum_{i=1}^{n+1}Z_i}\to 1 \ \ \ \text{ a.s. for } n\to\infty. $$ But then we have $$X_n=\frac{k_n}{n+1}\frac{\frac{1}{k_n}\sum_{i=1}^{k_n}Z_i} {\frac{1}{n+1}\sum_{i=1}^{n+1}Z_i}\to \nu \ \ \ \text{a.s., }$$ where we have used the fact that $$\lim_{n\to\infty} \frac{k_n}{n+1}=\nu$$