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 \(X_1,X_2,\ldots\) are independent and identically distributed, \(X_1^2,X_2^2,\ldots\) are also independent and identically distributed.
Thus, by the law of large numbers,
\[ Y_n=\frac{1}{n}\sum_{i=1}^{n}X_i^2 \] converges in probability to
\[ E(X_1^2) \]

Step 2: Compute \(E(X^2)\).
Since
\[ X\sim U(0,3), \] we have
\[ f(x)=\frac13,\qquad 0<x<3 \] Therefore,
\[ E(X^2)=\int_0^3 x^2\cdot \frac13\,dx \] \[ =\frac13\left[\frac{x^3}{3}\right]_0^3 \] \[ =\frac13\cdot 9 \] \[ =3 \]

Step 3: Final conclusion.
Hence,
\[ Y_n \xrightarrow{P} 3 \] Therefore, the required integer is
\[ \boxed{3} \]