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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?

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

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  • $\begingroup$ 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 $\endgroup$ – Xiaomi Oct 14 '18 at 9:32
  • $\begingroup$ @Xiaomi Good to hear that I could help! $\endgroup$ – Shashi Oct 14 '18 at 9:33

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