1. ## anyone help?

A random variable R has probability density,

p_R(x) = (x^m / m!)e^-x if x >= 0; and 0 otherwise

where m is a positive integer.

(a) Show that

P(0 <= R <= 2(m + 1)) = 1 - P(|R - (m + 1)| > m + 1)

(b) Use Chebyshev's inequality and part (a) to prove that

P(0 <= R <= 2(m + 1)) > m/m + 1

2. Originally Posted by bogazichili
A random variable R has probability density,

p_R(x) = (x^m / m!)e^-x if x >= 0; and 0 otherwise

where m is a positive integer.

(a) Show that

P(0 <= R <= 2(m + 1)) = 1 - P(|R - (m + 1)| > m + 1)

(b) Use Chebyshev's inequality and part (a) to prove that

P(0 <= R <= 2(m + 1)) > m/m + 1
(a) Note that $0 \leq R \leq 2(m + 1) \Rightarrow -(m + 1) \leq R - (m + 1) \leq m + 1 \Rightarrow | R - (m + 1) | \leq m + 1$.

(b) Note that $R$ follows a gamma distribution with $\alpha = m + 1$ and $\beta = 1$.

Therefore:

$\mu = \alpha \beta = m + 1$

$\sigma^2 = \alpha \beta^2 = m + 1 \Rightarrow \sigma = \sqrt{m + 1}$.

So $| R - (m + 1) | = | R - \mu |$ and $m + 1 = \sqrt{m + 1} \, \sigma$.

So apply Chebyshev's inequality to $\Pr( | R - \mu | \geq \sqrt{m + 1} \, \sigma)$.