Question

Let {X_n}_n≥1 be a sequence of independent and identically distributed random variables having the common probability density function
\(f(x) = \begin{cases} \frac{2}{x^3}, &  x \geq 1 \\     0, & \text{otherwise} . \end{cases}\)
If \(\lim\limits_{n \rightarrow \infin}P(|\frac{1}{n}\sum^n_{i=1}X_i-\theta|< \in)=1\) for all ∈ > 0, then θ equals

• A. 4
• B. 2
• C. ln 4
• D. ln 2

Answer 1

The Correct Option is B

To solve the given problem, we need to understand the expression provided: \(\lim\limits_{n \rightarrow \infty}P\left(\left|\frac{1}{n}\sum^n_{i=1}X_i-\theta\right|< \in\right)=1\) for all \(\epsilon > 0\). This expression resembles the concept of the Law of Large Numbers, which suggests that the sample mean converges to the expected value of the distribution as the number of samples approaches infinity.

The sequence \({X_n}\) consists of independent and identically distributed (i.i.d.) random variables with the given probability density function:

\[f(x) = \begin{cases} \frac{2}{x^3}, & x \geq 1 \\ 0, & \text{otherwise} \end{cases}\]

First, we find the expected value \(E(X)\) of one random variable \(X\) with the given density function:

\[E(X) = \int_{1}^{\infty} x \cdot \frac{2}{x^3} \,dx = \int_{1}^{\infty} \frac{2}{x^2} \,dx\]

To solve the integral:

\[\int \frac{2}{x^2} \, dx = 2 \int x^{-2} \, dx = 2 \left[ \frac{x^{-1}}{-1} \right] = -\frac{2}{x}\]

Evaluating it from 1 to \(\infty\):

\[E(X) = \left[ -\frac{2}{x} \right]_{1}^{\infty} = \left(0 - (-2)\right) = 2\]

Thus, the expected value \(E(X)\) is 2. By the Law of Large Numbers, the sample mean converges to the expected value. Therefore, \(\theta\), the value it converges to, is the expected value:

Therefore, \(\theta = 2\).

Hence, the correct answer is the option '2'.