# Slutsky's theorem applied to Bernoulli random variables

Suppose $$X_n$$ is a sequence of $$n$$ Bernoulli random variables with unknown $$p$$, and I try to get a confidence interval for $$p$$. Using central limit theorem, I've got

\begin{align*} \frac{\bar X_n - p}{\sqrt{(p(1-p)/n}} \sim \mathcal{N}(0, 1) \\ \end{align*}

Suppose

$$\mathbb{P}\Bigg(\left| \frac{\bar X_n - p}{\sqrt{(p(1-p)/n}} \right|\Bigg) \le t$$

then we get a confidence interval for $$p$$ with a probability of $$t$$:

$$\mathcal{I}_{t} = \Big[\bar X_n - \frac{t\sqrt{p(1-p)}}{\sqrt{n}}, \bar X_n + \frac{t\sqrt{p(1-p)}}{\sqrt{n}} \Big]$$

Unfortunately, this interval depends on $$p$$. I've learned that somehow we could replace $$p$$ with $$\bar X_n$$ when $$n \to \infty$$, and the interval becomes

$$\mathcal{I}_t' = \Big[\bar X_n - \frac{t\sqrt{\bar X_n(1-\bar X_n)}}{\sqrt{n}}, \bar X_n + \frac{t\sqrt{\bar X_n(1-\bar X_n)}}{\sqrt{n}} \Big]$$

I feel it may have something related to Slutsky's theorem, but I don't how it could be applied in this case. Could anybody explain with some more details, please?

Slutsky's theorem says that if $$X_n \xrightarrow[n \to \infty]{(d)} X$$ and $$Y_n \xrightarrow[n \to \infty]{\mathbb{P}} c$$, then

\begin{align*} X_n Y_n &\xrightarrow[n \to \infty]{(d)} cX \\ X_n + Y_n &\xrightarrow[n \to \infty]{(d)} X + c \end{align*}

• $$\xrightarrow[n \to \infty]{(d)}$$ means convergence in distribution
• $$\xrightarrow[n \to \infty]{\mathbb{P}}$$ means convergence in probability

NOTE: if $$X$$ in the above example is also a constant, the result is strengthened to convergence in probability.

In your example we have $$\bar{X} \xrightarrow[n \to \infty]{\mathbb{P}} p$$ by the Weak Law of Large Numbers. Hence, if you can show that

$$\sqrt{\frac{\bar{X_n}(1-\bar{X_n})}{n}} \xrightarrow[n \to \infty]{(d)} \sqrt{\frac{p(1-p)}{n}}$$

then by applying Slutsky's theorem will tell you that their sum converges in probability:

$$\bar{X_n} \pm t\sqrt{\frac{\bar{X_n}(1-\bar{X_n})}{n}} \xrightarrow[n \to \infty]{\mathbb{P}} \bar{X_n} \pm t\sqrt{\frac{p(1-p)}{n}}$$

To prove $$\sqrt{\frac{\bar{X_n}(1-\bar{X_n})}{n}} \xrightarrow[n \to \infty]{(d)} \sqrt{\frac{p(1-p)}{n}}$$, note that by the central limit theorem we have

$$\sqrt{n}(\bar{X_n} - p) \xrightarrow[n \to \infty]{(d)} N(0,p(1-p))$$

Since $$\bar{X_n}$$ is the average of i.i.d $$\rm{Bernoulli}(p)$$ random variables. Let $$f(x) = \sqrt{x(1-x)}$$ which is continuous on the support of $$X_n$$ (since $$X_n \geq 0)$$. So we can apply the Delta method to obtain

$$\sqrt{n}\left(\sqrt{\bar{X_n}(1-\bar{X_n})} - \sqrt{p(1-p)}\right) \xrightarrow[n \to \infty]{(d)} N(0,f'(p)^2 p(1-p))$$

That is,

$$\sqrt{\bar{X_n}(1-\bar{X_n})} \xrightarrow[n \to \infty]{(d)} N\left(\sqrt{p(1-p)}, \frac{f'(p)^2 p(1-p)}{n}\right)$$

Notice the variance's numerator only depends on $$p$$, so it's fixed, but the denominator depends on $$n$$. So as $$n\to \infty$$ the variance goes to $$0$$, and hence

$$\sqrt{\bar{X_n}(1-\bar{X_n})} \xrightarrow[n \to \infty]{\mathbb{P}} \sqrt{p(1-p)}$$

It immediately follows that

$$\sqrt{\frac{\bar{X_n}(1-\bar{X_n})}{n}} \xrightarrow[n \to \infty]{\mathbb{P}} \sqrt{\frac{p(1-p)}{n}}$$

and hence distribution, as required.

• That's insightful! But how to show $\sqrt{\frac{\bar{X_n}(1-\bar{X_n})}{n}} \to \sqrt{\frac{p(1-p)}{n}}$? I almost took it for granted, as the convergence in the numerator obvious, but the denominator also goes to $\infty$. Is there a rigorous way to prove it, please? – zyxue Oct 4 '18 at 4:21
• Added the proof at the end of my comment. This stuff becomes easy once you've done a few proofs using CLT/Slutsky's theorem/Delta method – Xiaomi Oct 4 '18 at 4:41
• Is it true that if $\bar X_n \to p$, then $\sqrt{\bar X_n} \to \sqrt{p}$? – zyxue Oct 8 '18 at 20:42
• Yes, by the continuous mapping theorem since $\bar{X_n}$ is non-negative. – Xiaomi Oct 8 '18 at 23:19