Answer to Question #85925 in Statistics and Probability for Umalaxmi

Question

For the random variable X with density function f(x)={2e^(-2x) ; x≥0
{0 ;x<0.
Find (i)the mean μ and variance σ^(2);
(ii)P [|X-μ|≥1].
Use Chebyshev's inequality to obtain an upper bound on P[|X-μ|]≥1] and compare with the result obtained in part (ii).

Answer 1

$$\mu=\int\limits_{\mathbb{R}} xf(x) \, dx=\int\limits_0^{\infty} 2xe^{-2x}\,dx=0.5$$

$$\sigma^2=\int\limits_{\mathbb{R}} (x-\mu)^2 f(x) \, dx=\int\limits_0^{\infty} 2(x-0.5)^2 e^{-2x}\,dx=0.25$$

$$P(|X-\mu| \geqslant 1=2 \sigma)=\int\limits_{|x-0.5|\geqslant 1} f(x) \, dx=\int\limits_{1.5}^{\infty} 2e^{-2x}\,dx=e^{-3}$$

$$P(|X-\mu| \geqslant 1=2 \sigma) \leqslant \dfrac{1}{2^2}=0.25$$

$$e^{-3}\approx 0.05<0.25$$