Bound for the exponential distribution using chebychev's inequality

Question

Let $X$ be an exponential distribution with rate lambda. I'm supposed to find the exact value of $P(|X-\mu |\ge k \sigma )$ for any $k>1$ then compare it to the bound I get from Chebyshev's. How do I use Chebyshev's to find a bound when I'm not given anything else? I'm pretty confused on Chebyshev unless I'm given numbers.

Answer 1

Our random variable $X$ has density function $\lambda e^{-\lambda x}$ for $x\gt 0$. Note that $X$ has mean $\mu=\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^2}$.

For $k\ge 1$, the probability that $|X-\mu|$ is $\ge \frac{k}{\lambda}$ is $e^{-(\lambda)(k/\lambda)}$, which is $e^{-k}$. This is because $\frac{1}{\lambda}-\frac{k}{\lambda}\le 0$. (For $0\lt k\lt 1$, the expression for the probability would be different.)

The Chebyshev Inequality says that the probability that $|X-\mu|\ge k\sigma$, that is, $\frac{k}{\lambda}$, is $\le \frac{1}{k^2}$.

Note that the bound given by the Chebyshev Inequality is some distance from the true probability.