Question

If \(X\) is a random variable with the cumulative distribution function

• A. \(\sqrt{\dfrac{\pi}{2}}\)
• B. \(\sqrt{2\pi}\)
• C. \(\sqrt{\dfrac{2}{\pi}}\)
• D. \(1\)

Hint:

If the CDF is given, first differentiate it to obtain the PDF, and then use the definition of expectation.

Answer 1

The Correct Option is A

Step 1: Find the probability density function.
The cumulative distribution function is
\[ F(x)= \begin{cases} 0, & x\leq 0,\\ 1-e^{-\frac{x^2}{2}}, & x>0. \end{cases} \] Differentiate \(F(x)\) for \(x>0\):
\[ f(x)=\frac{d}{dx}\left(1-e^{-\frac{x^2}{2}}\right) \] Using chain rule,
\[ f(x)=xe^{-\frac{x^2}{2}}, \qquad x>0 \] Hence, the probability density function is
\[ f(x)= \begin{cases} xe^{-\frac{x^2}{2}}, & x>0,\\ 0, & x\leq 0. \end{cases} \]

Step 2: Write the expression for expectation.
The expectation is
\[ E(X)=\int_{0}^{\infty}x f(x)\,dx \] Substituting \(f(x)\),
\[ E(X)=\int_{0}^{\infty}x\left(xe^{-\frac{x^2}{2}}\right)dx \] \[ =\int_{0}^{\infty}x^2e^{-\frac{x^2}{2}}dx \]

Step 3: Evaluate the integral using Gamma-function formula.
We use the standard result
\[ \int_{0}^{\infty}x^ne^{-ax^2}dx = \frac{1}{2}a^{-\frac{n+1}{2}} \Gamma\left(\frac{n+1}{2}\right) \] Here,
\[ n=2,\qquad a=\frac12 \] Therefore,
\[ E(X) = \frac12\left(\frac12\right)^{-\frac32} \Gamma\left(\frac32\right) \] Now,
\[ \left(\frac12\right)^{-\frac32}=2^{\frac32}=2\sqrt2 \] and
\[ \Gamma\left(\frac32\right) = \frac12\sqrt{\pi} \] Thus,
\[ E(X) = \frac12(2\sqrt2)\left(\frac12\sqrt{\pi}\right) \] \[ = \frac{\sqrt{2\pi}}{2} \] \[ = \sqrt{\frac{\pi}{2}} \]

Step 4: Final conclusion.
Hence,
\[ \boxed{E(X)=\sqrt{\frac{\pi}{2}}} \] Therefore, the correct option is
\[ \boxed{(A)} \]