Sample Mean of a Product Sequence
You have two independent sequences of IID random variables: $X_1, X_2, \ldots$ with mean $\mu_X$ and variance $\sigma_X^2$, and $Y_1, Y_2, \ldots$ with mean $\mu_Y$ and variance $\sigma_Y^2$. The two sequences are independent of each other. Define $Z_i = X_i Y_i$.
Does the sample mean
$$\frac{Z_1 + Z_2 + \cdots + Z_n}{n}$$
converge to a limiting distribution as $n \to \infty$? If it does, state the sum of the mean and variance of that limiting distribution. If it does not converge, answer $-100$.
Use $\mu_X = 5$, $\mu_Y = 6$, $\sigma_X^2 = 25$, $\sigma_Y^2 = 36$.
Hints
- The $Z_i = X_i Y_i$ are IID with finite mean. Think about which limit theorem applies to a sample mean of IID terms -- not the CLT, which requires standardization.
- The Weak Law of Large Numbers says the sample mean of IID variables with finite mean converges in probability to $E[Z_1]$. That is a constant, not a distribution with nonzero variance.
- Compute $E[X_1 Y_1]$ using the independence of the $X$ and $Y$ sequences: $E[X_1 Y_1] = E[X_1] \cdot E[Y_1] = \mu_X \mu_Y$. Then evaluate at the given values.
Worked Solution
How to Think About It: The first question an interviewer is probing here is: do you know when to reach for the LLN versus the CLT? The sample mean of $n$ IID terms -- that is the LLN setup, not CLT. The CLT applies when you standardize the sum; here we are just dividing by $n$, so the sum concentrates on its expectation. Before worrying about the specific values of $\mu$ and $\sigma^2$, check: are the $Z_i$ IID with finite mean? Yes -- products of independent IID sequences are themselves IID, and finite moments of $X$ and $Y$ give finite moments of $XY$. So by the LLN, the sample mean converges to $E[Z_1]$. That is a constant -- no distribution, just a point mass. The "variance of the limiting distribution" is zero.
Quick Estimate: Before the formal work, sanity-check the answer. With $\mu_X = 5$ and $\mu_Y = 6$, and since $X_i$ and $Y_i$ are independent, we expect $E[X_i Y_i] = E[X_i] \cdot E[Y_i] = 5 \times 6 = 30$. So the sample mean should converge to $30$. The limiting "distribution" is a point mass at $30$, with mean $30$ and variance $0$. The answer is $30 + 0 = 30$.
Approach: Formally verify that $Z_i$ is IID with finite mean, apply the Weak Law of Large Numbers, and compute $E[Z_1]$ using independence.
Formal Solution:
First, check the conditions for the LLN:
- Each $Z_i = X_i Y_i$ is a measurable function of $(X_i, Y_i)$. Since the $(X_i, Y_i)$ pairs are independent across $i$ (by independence of the two sequences) and identically distributed, the $Z_i$ are IID.
- $E[|Z_1|] = E[|X_1 Y_1|] = E[|X_1|] \cdot E[|Y_1|]$ by independence. Finite variance implies finite mean, so both factors are finite. Thus $E[|Z_1|] < \infty$.
By the Weak Law of Large Numbers:
$$\frac{Z_1 + Z_2 + \cdots + Z_n}{n} \xrightarrow{p} E[Z_1]$$
Now compute $E[Z_1]$:
$$E[Z_1] = E[X_1 Y_1] = E[X_1] \cdot E[Y_1] = \mu_X \cdot \mu_Y$$
where the factorization uses independence of $X_1$ and $Y_1$.
The limiting "distribution" is a point mass at $\mu_X \mu_Y = 5 \times 6 = 30$. Its mean is $30$ and its variance is $0$.
$$\text{Mean} + \text{Variance} = 30 + 0 = 30$$
Answer: $\boxed{30}$
Intuition
The key distinction here is LLN vs. CLT. Both involve sums of IID terms, but they answer different questions. The CLT tells you about the distribution of the standardized sum $(S_n - n\mu) / (\sigma \sqrt{n})$ -- it is always approximately Normal. The LLN tells you about the raw sample mean $S_n / n$ -- it collapses to a point mass at $\mu$. Dividing by $n$ without subtracting the mean and rescaling kills all the randomness in the limit. This trips people up because they reflexively reach for the CLT whenever they see a sum. The interview is testing whether you know when NOT to use it.
The second lesson is how independence interacts with expectation. A lot of candidates know $E[XY] = E[X] E[Y]$ for independent variables, but hesitate when the independence is "cross-sequence" rather than from a single joint distribution. Here, $X_i$ and $Y_i$ are independent because they come from two entirely separate sequences -- that is enough for the product rule to apply. In practice, this type of reasoning shows up whenever you model returns or signals as products of independent factors (e.g., a factor model where magnitude and sign are modeled separately), and you need to quickly compute the expected value of the product.