# Limit of a probability involving a sum of independent binary random variables

The independent binary random variables $$X_k$$ take values $$\pm 1$$ with the probabilities $$(1 \pm k^{\frac{-1}{2}})/2, k = 1,2,...$$. Find $$\lim_{n \rightarrow \infty} \mathbb{P}(X_1 + ... + X_n \leq 0)$$.

Here are my thoughts so far:

I'd like to use the Central Limit Theorem here -- if $$X_1,...,X_n$$ are a sequence of i.i.d. random variables, each with mean $$\mu$$ and variance $$\sigma^2$$, then the distribution of $$\frac{X_1 + ... + X_n - n\mu}{\sigma \sqrt{n}}$$ tends to the standard normal as $$n \rightarrow \infty$$.

However, here, although $$X_1,..,X_n$$ are independent, they are not identically distributed.

I first let $$Y_k = \frac{X_k + 1}{2}$$, so that $$P(Y _k = 0) = \frac{1 - k^{\frac{-1}{2}}}{2}$$ and $$P(Y _k = 1) = \frac{1 + k^{\frac{-1}{2}}}{2}$$. This gives that $$Y_k$$ is a Bernoulli random variable with probability of success $$\frac{1 + k^{\frac{-1}{2}}}{2}$$. This makes our calculations easier to handle, since we're now working with random variables we're familiar with -- but still, $$Y_1,...,Y_n$$ are independent but not identically distributed.

Where can I continue from here ? Another idea is to compute the moment generating function of $$X_1 + ... + X_n$$, and map this back to the p.d.f. of a familiar random variable by the uniqueness theorem -- but, the resulting moment generating function doesn't map to such a familiar p.d.f., as the sum of Bernoulli random variables with different success probabilities does not give a nice, familiar distribution. Thus, I'm sticking with my instinct that the Central Limit Theorem is the way to go here.

Any help would be appreciated. Thanks!

• I computed $$\mathbb E[e^{tX_k}] = \frac12\left((1+k^{-1/2})e^t - (1-k^{-1/2})e^{-t} \right)$$ and hence $$\mathbb E[e^{tS_n}] = \prod_{k=1}^n \mathbb E[e^{tX_1}] = \frac12 \prod_{k=1}^n ((1+k^{-1/2}e^t - (1-k^{-1/2})e^{-t}))$$ which doesn't seem to have a nice closed form. Jan 13, 2020 at 23:10
• Binary random variables are usually called Bernoulli.
– Mark
Jan 13, 2020 at 23:45

The random variables $$\ X_n\$$ satisfy the Lindeberg condition, and hence the central limit theorem holds, in the form $$\frac{\sum_{k=1}^n\left(X_k-\mathbb{E}\left(X_k\right)\right)}{s_n}\overset{\mathcal D}{\rightarrow}\mathcal{N}(0,1)\ \ \text{ as }\ n\rightarrow\infty\ ,$$ where $$\ s_n=\sqrt{\sum_\limits{k=1}^n \mathbb{E} \left(\left(X_k-\mathbb{E}\left(X_k\right)\right)^2 \right)}\$$.
Here, $$\mathbb{E}\left(X_n\right)=\frac{1}{\sqrt{n}}\ \ \text{and}\\ s_n^2=n-\sum_{k=1}^n\frac{1}{k}\ ,$$ so $$\lim_\limits{n\rightarrow\infty} \frac{\sum_{k=1}^n \mathbb{E}\left(X_k\right)}{s_n} = \lim_\limits{n\rightarrow\infty}\frac{\sum_{k=1}^n \frac{1}{\sqrt{k}}}{s_n}=2\ ,$$ and \begin{align} \lim_\limits{n\rightarrow\infty}\mathbb{P}\left(\sum_{k=1}^nX_k\le0\right)&= \lim_\limits{n\rightarrow\infty}\mathbb{P}\left(\frac{\sum_{k=1}^n\left(X_k-\mathbb{E}\left(X_k\right)\right)}{s_n}\le-\frac{\sum_{k=1}^n \frac{1}{\sqrt{k}}}{s_n}\right)\\ &= \mathcal{N}\left(0,1;-2\right)\\ &\approx0.023\ . \end{align} That $$\ X_n\$$ satisfy the Lindeberg condition is not obvious, but unless there's a blunder in my arithmetic (of which the probability is not entirely negligible) the proof is fairly straightforward. I'll post it here if there's a request for me to do so.
• Minor comment: $\sum_1^n\frac{1}{\sqrt{k}}\sim 2\sqrt{n}$.