Computing a limit of a probability using CLT

Question

Suppose $X\sim\mathsf{Bin}\left({18},\frac13\right)$ and $Y\sim\mathsf{Bin}\left({m},\frac13\right)$ independent random variables, compute: $$\displaystyle{ \lim_{m\to\infty}\mathbb{P}\left[X\leq t-Y\right] }$$while $t=\frac{m}{4}.$

I've set $Z=X+Y$ and then said that $Z$ is a sum of $m+18$ independent bernouli rv's with $p=\frac{1}{3}$ but then I've tried using the central limit theorem but since $t=\frac{m}{4}$ I cant get the thing inside the probability to look like $\frac{S_n-n\mu}{\sigma\sqrt{n}}\leq k$ for some $k$.

Answer 1

You have $$ \mathbb P(X+Y \leq t) = \mathbb P(Z_m \leq t) $$ where $Z_m = \mbox{Bin}(18+m,\frac 13)$ (in distribution) is a sum of $18+m$ independent Bernoulli random variables with parameter $p=\frac 13$.

In particular, $\mathbb EZ_m = 6+\frac{m}{3}$ and $\mbox{Var}(Z_m) = (18+m)\times \frac{1}{3} \times\frac{2}{3} = 4+\frac{2m}{9}$. Therefore, $$ \frac{Z_m - \mathbb EZ_m}{\sqrt{\mbox{Var}(Z_m)}} = \frac{Z_m - 6-\frac{m}{3}}{\sqrt{4+\frac{2m}{9}}} \to N(0,1) $$ in distribution, by the central limit theorem.

---

Let's do some "unrigorous" arithmetic. Roughly speaking, $$Z_m - 6-\frac{m}{3} \approx \sqrt{4+\frac{2m}{9}}N(0,1)= N\left(0,4+\frac{2m}{9}\right) \implies Z_m \approx N\left(6+\frac{m}{3},4+\frac{2m}{9}\right)$$

We want $\mathbb P(Z_m \leq \frac{m}{4})$, which we rewrite as $\mathbb P(\frac{4Z_m}{m} \leq 1)$. From the heuristics above, $$ Z_m \approx N\left(6+\frac{m}{3},4+\frac{2m}{9}\right) \implies \frac{4Z_m}{m} \approx N\left(\left(6+\frac{m}{3}\right)\times \frac{4}{m},\left(4+\frac{2m}{9}\right) \times \frac{16}{m^2}\right) $$ As $m \to \infty$, we observe that the mean converges to $\frac 43$ : but the variance converges to $0$. Hence, we are led to believe that $\frac{4Z_m}m \to \frac{4}{3}$, the constant random variable, in distribution. Can we prove this rigorously now? Obviously, the arguments above are not rigorous because the $\approx$ signs do not indicate any equality, but rather comparisons in some appropriate asymptotic sense that we have not quantified.

---

All this can be achieved by some simple manipulations. First, we multiply both top and bottom of the CLT left hand side by $\frac{4}{m}$. This leads to $$ \frac{\frac{4Z_m}{m} - \frac{24}{m} - \frac 43}{\frac{4}{m}\sqrt{4+\frac{2m}{9}}} \to N(0,1). $$ Observe that the sequence $\frac{4}{m}\sqrt{4+\frac{2m}{9}} \to 0$ as $m \to \infty$. Treating this as a sequence of constant random variables which are converging in probability to a constant and applying Slutsky's theorem(see the end of this answer), $$ \frac{\frac{4Z_m}{m} - \frac{24}{m} - \frac 43}{\frac{4}{m}\sqrt{4+\frac{2m}{9}}} \times \left(\frac{4}{m}\sqrt{4+\frac{2m}{9}}\right) \to N(0,1) \times 0 = 0 $$ in distribution. This is equivalent to $\frac{4Z_m}{m} - \frac{24}{m} - \frac 43 \to 0$ in distribution. However, we may apply Slutsky's theorem yet again. This time, the sequence of constant random variables $\frac{24}{m}+ \frac 43 \to \frac{4}{3}$ in probability. Hence, applying Slutsky's theorem $$ \frac{4Z_m}{m} - \frac{24}{m} - \frac 43 +\frac{24}{m}+ \frac 43 \to 0+\frac{4}{3} = \frac{4}{3}. $$ In particular, we have proved that $\frac{4Z_m}{m} \to \frac{4}{3}$ in distribution. From here, observe that the CDF of the constant random variable $\frac{4}{3}$ is continuous at the point $1$, therefore $$ \mathbb P\left(\frac{4Z_m}{m} \leq 1\right) \to \mathbb P\left(\frac 43 \leq 1\right) = 0, $$ concluding the proof.

---

Slutsky's theorem : If $X_n \to X$ in distribution and $Y_n \to Y$ in probability, where $Y$ is a (finite) constant random variable, then $X_nY_n \to XY$ and $X_n+Y_n \to X+Y$ in distribution. We used both versions above.

Answer 2

Remember $\mu=1/3$ in this case and that $\sigma=\sqrt{2/9}$. So the z-scores are: $$Z_m=\frac{3(X+Y)-(m+18)}{\sqrt{2(m+18)}}$$The probability $X+Y<t$ is the probability that: $$Z_m<\frac{3m/4-(m+18)}{\sqrt{2(m+18)}}$$Can you see why this should tend to zero as $m\to\infty$? Try bounding the right hand side by a fixed number (which you make arbitrarily small...) and taking limits in the left hand side.

Answer 3

You are essentially asking for $${ \lim\limits_{m\to\infty}\mathbb{P}\left[Z_m\leq \frac m4\right] }$$

where $Z_m\sim\mathsf{Bin}\left({18+m},\frac13\right)$

• let $k=18+m$ and $W_k=W_{18+m}=Z_m$ with $W_k \sim \mathsf{Bin}\left({k},\frac13\right)$, i.e. the sum of $k$ iid Bernoulli random variables with $p=\frac13$
• so the question becomes $${ \lim\limits_{k\to\infty}\mathbb{P}\left[W_k\leq \frac {k-18}4\right] }$$
• and this is bounded above by ${ \lim\limits_{k\to\infty}\mathbb{P}\left[W_k\leq \frac {k}4\right] }={ \lim\limits_{k\to\infty}\mathbb{P}\left[\frac1k W_k\leq \frac {1}4\right] }$
• the laws of large numbers tell us $\frac1k W_k \to \frac13$ in probability and almost surely (you can also use the CLT, though it is not needed)
• and since $\frac14 < \frac13$ we have ${ \lim\limits_{k\to\infty}\mathbb{P}\left[\frac1k W_k\leq \frac {1}4\right] }=0$
• meaning this is an upper bound on ${ \lim\limits_{m\to\infty}\mathbb{P}\left[Z_m\leq \frac m4\right] }$ and on ${ \lim\limits_{m\to\infty}\mathbb{P}\left[X\leq t-Y\right] }$
• so they too are $0$.