Question

Let \(\{X_k\}_{k\geq1}\) be a sequence of independent random variables such that \[ X_{2k-1}\sim \text{Bin}(1,\theta),\quad\text{and}\quad X_{2k}\sim \text{Bin}(1,1-\theta),\quad k=1,2,3,\ldots, \] where \(\theta\in(0,1)\). Let \(\{Y_k\}_{k\geq1}\) be another sequence of independent and identically distributed random variables such that \(Y_k\sim\text{Poisson}(\lambda)\), \(\lambda>0\). Define, for \(n\in\mathbb{N}\), \[ S_{2n}=\sum_{k=1}^{n}(X_{2k-1}-X_{2k}+1-2\theta),\quad W_n=\sum_{k=1}^{n}Y_k^2\quad\text{and}\quad \sigma_{2n}^2=2n\theta(1-\theta). \] Then which of the following statements is/are correct?

• A. \(\dfrac{nS_{2n}}{\sigma_{2n}W_n}\xrightarrow{d}N(0,1),\) as \(n\to\infty\)
• B. \(\dfrac{nS_{2n}}{\sigma_{2n}W_n}\xrightarrow{d}N\!\left(0,\ \dfrac{1}{\lambda^2(1+\lambda)^2}\right),\) as \(n\to\infty\)
• C. \(\dfrac{nS_{2n}+\sigma_{2n}W_n}{n\sigma_{2n}}\xrightarrow{d}N(\lambda(1+\lambda),\ 1),\) as \(n\to\infty\)
• D. \(\dfrac{nS_{2n}+\sigma_{2n}W_n}{n\sigma_{2n}}\xrightarrow{d}N\!\left(0,\ \dfrac{1}{\lambda^2(1+\lambda)^2}\right),\) as \(n\to\infty\)

Hint:

Show S_2n/sigma_2n converges to N(0,1) by the Central Limit Theorem and W_n/n converges in probability to lambda(1+lambda) by the Law of Large Numbers, then combine the two limits using Slutsky's theorem.

Answer 1

The Correct Option is B, C

Step 1: Simplify S_2n/sigma_2n as a standardized sum.
Write $U_k=X_{2k-1}-X_{2k}+1-2\theta$. Each $U_k$ has mean 0 and, using independence of the two Bernoulli pieces, variance $2\theta(1-\theta)$. Since the $U_k$ are iid across k, the classical CLT gives
\[ \frac{\sum_{k=1}^n U_k}{\sqrt{2n\theta(1-\theta)}}=\frac{S_{2n}}{\sigma_{2n}}\xrightarrow{d}Z, \]
where $Z\sim N(0,1)$.

Step 2: Simplify W_n/n as a sample mean.
$W_n/n=\frac{1}{n}\sum_{k=1}^n Y_k^2$ is the sample average of iid variables $Y_k^2$ with $E(Y_k^2)=\lambda(1+\lambda)$ and finite variance, since all Poisson moments are finite. The Weak Law of Large Numbers gives $W_n/n\to\lambda(1+\lambda)$ in probability; call this constant $c=\lambda(1+\lambda)$.

Step 3: Handle the ratio nS_2n/(sigma_2n W_n).
This equals $Z_n\cdot\frac{1}{(W_n/n)}$ where $Z_n=S_{2n}/\sigma_{2n}\to Z\sim N(0,1)$ and $\frac{1}{W_n/n}\to\frac{1}{c}$ in probability. By Slutsky's theorem, a distributional limit times a probability limit gives $\frac{1}{c}Z\sim N\left(0,\frac{1}{c^2}\right)=N\left(0,\frac{1}{\lambda^2(1+\lambda)^2}\right)$. This is option (B), so it is TRUE, while (A) fails since the variance is not 1 in general.

Step 4: Handle the sum form.
$\frac{nS_{2n}+\sigma_{2n}W_n}{n\sigma_{2n}}=Z_n+\frac{W_n}{n}\to Z+c$ in distribution, again by Slutsky, since adding a constant probability limit to a distributional limit only shifts the mean. So the limit is $N(c,1)=N(\lambda(1+\lambda),1)$, matching (C), while (D) with mean 0 is wrong.

Final Answer:
Slutsky's theorem gives N(0, 1/(lambda squared (1+lambda) squared)) for the ratio and N(lambda(1+lambda), 1) for the sum. \[ \boxed{\text{(B) and (C)}} \]