# Almost sure convergence of a beta distribution random variable

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?

• How does the limiting random variable still depend on $n$ after taking $n$ to infinity? – Shashi Oct 14 '18 at 8:50
• Sorry, I should have specified, $k$ also increases with $n$. It is defined as $k = [n \nu ]$ for some $\nu \in (0,1)$. So $k/n \to \nu$. – Xiaomi Oct 14 '18 at 8:54
• I would suggest you write $X_n\to \nu$ in that case... BTW what are these square brackets? Floor function? – Shashi Oct 14 '18 at 9:02
• Yes, floor function/integer part. – Xiaomi Oct 14 '18 at 9:05

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 $$\lim_{n\to\infty} \frac{k_n}{n+1}=\nu$$
• Thank you, especially for setting me straight on notation. I realise now it's non-sensical to write $lim X_n = k/n$. Will make sure to keep that in mind in future – Xiaomi Oct 14 '18 at 9:32