Question

For the random variable X with density function
f(x)=[2e^-2x ; x≥0
[0 ; <0
Find
i) the mean μand variance σ²;
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

$\mu=\int_{0}^{\infty}2xe^{-2x}dx=1/2\\ \sigma^{2}=\int_0^{\infty}2(x-1/2)^{2}e^{-2x}dx=1/4\\ P(|X-\mu|>=1)=1-P(|X-\mu|<1)=1-\int_{0}^{3/2}2e^{-2x}dx=e^{-3}=0.0498\\$

By Chebyshev's inequality: $P(|X-\mu|>=1)=P(|X-\mu|>=2\sigma)<=1/4=0.25.$