Question

Let \(\{X_n\}_{n\geq 1}\) be a sequence of independent and identically distributed random variables having \(U(0,3)\) distribution. If \(Y_n=\frac{1}{n}\sum_{i=1}^{n}X_i^2,\; n\geq 1\), then \(\{Y_n\}_{n\geq 1}\) converges in probability to (in integer).

Hint:

By the law of large numbers, the sample average of \(g(X_i)\) converges in probability to \(E[g(X)]\), provided the expectation exists.

Answer 1

Correct Answer: 3

Step 1: Apply the law of large numbers.
Since the $X_i$ are iid, so are the $X_i^2$, so $Y_n=\frac1n\sum X_i^2$ converges in probability to $E(X_1^2)$.

Step 2: Set up $E(X^2)$.
For $X\sim U(0,3)$ the density is $\frac13$ on $(0,3)$.

Step 3: Integrate.
$E(X^2)=\int_0^3 x^2\cdot\frac13 dx=\frac13\cdot\frac{27}3=\frac13\cdot9=3$.

Step 4: State the limit.
So $Y_n\xrightarrow{P}3$.

Step 5: Conclude.
\[ \boxed{3} \]