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: Study one term of S_2n.
Let \(T_k=X_{2k-1}-X_{2k}+1-2\theta\). Since \(X_{2k-1}\sim\text{Bin}(1,\theta)\) has mean \(\theta\) and \(X_{2k}\sim\text{Bin}(1,1-\theta)\) has mean \(1-\theta\),
\[ E(T_k)=\theta-(1-\theta)+1-2\theta=0. \]
Both Bernoulli pieces have variance \(\theta(1-\theta)\) (the same number either way), and they are independent, so
\[ \text{Var}(T_k)=\theta(1-\theta)+\theta(1-\theta)=2\theta(1-\theta). \]

Step 2: Apply the Central Limit Theorem to S_2n. \(S_{2n}=\sum_{k=1}^{n}T_k\) is a sum of n iid mean-zero terms with variance \(2\theta(1-\theta)\), so \(\text{Var}(S_{2n})=2n\theta(1-\theta)=\sigma_{2n}^2\), matching the given notation. So
\[ \frac{S_{2n}}{\sigma_{2n}}\xrightarrow{d}N(0,1). \]

Step 3: Study W_n/n.
For \(Y_k\sim\text{Poisson}(\lambda)\), \(E(Y_k^2)=\text{Var}(Y_k)+(EY_k)^2=\lambda+\lambda^2=\lambda(1+\lambda)\). By the Weak Law of Large Numbers,
\[ \frac{W_n}{n}\xrightarrow{P}\lambda(1+\lambda). \]

Step 4: Combine using Slutsky's theorem for (A) and (B).
\[ \frac{nS_{2n}}{\sigma_{2n}W_n}=\frac{S_{2n}}{\sigma_{2n}}\cdot\frac{n}{W_n}. \]
Since \(\frac{n}{W_n}\xrightarrow{P}\frac{1}{\lambda(1+\lambda)}\), Slutsky's theorem gives
\[ \frac{nS_{2n}}{\sigma_{2n}W_n}\xrightarrow{d}N\!\left(0,\ \frac{1}{\lambda^2(1+\lambda)^2}\right). \]
So (B) is TRUE and (A) is FALSE.

Step 5: Combine for (C) and (D).
\[ \frac{nS_{2n}+\sigma_{2n}W_n}{n\sigma_{2n}}=\frac{S_{2n}}{\sigma_{2n}}+\frac{W_n}{n}. \]
Adding a term converging in distribution to N(0,1) to a term converging in probability to the constant \(\lambda(1+\lambda)\), Slutsky's theorem gives
\[ \xrightarrow{d}N(\lambda(1+\lambda),\ 1). \]
So (C) is TRUE and (D) is FALSE.

Final Answer:
The ratio form converges to N(0, 1/(lambda squared times (1+lambda) squared)) and the sum form converges to N(lambda(1+lambda), 1). \[ \boxed{\text{(B) and (C)}} \]