Given a positive valued random variable with known mean and variance, find the probability of the random variable being greater than 5

Question

I have a positive valued random variable X with mean (expected value) 2 and variance 1. I need to prove that the probability of X being greater than 5 is less than or equal to $\frac{1}{5}$.

Given: $$X > 0, \mu = 2, \sigma^2 = 1$$ Prove: $$P(X > 5) ≤ \frac{1}{5}$$

I know nothing of how X is distributed or even if X is a discrete or continuous random variable. The information above is everything given in the problem statement.

The one hint that I have received is to make use of this corollary "monotone transformation":

$$P(|X| ≥ a) ≤ P(\phi(X) ≥ \phi(a))≤\frac{E[\phi(X)]}{\phi(a)}$$

and "carefully design the function $\phi$".

I know that the middle and right parts of the corollary basically represent Markov's inequality, but beyond that I don't understand what this means.

I could use Markov's inequality alone to show that the probability of X being greater than 5 is less than or equal to $\frac{2}{5}$, but that is not good enough. How do I use the above hint to prove the probability is smaller than $\frac{1}{5}$?

Answer 1

We have $$ P(X > 5) = P(X^2 > 25) \leq P(X^2 \geq 25) \leq \frac{\mathsf{E}(X^2)}{25} $$ Moreover, since $\mathsf{E}(X) = 2$ and $\mathsf{Var}(X) = 1$, $$ \mathsf{E}(X^2) = \mathsf{E}(X)^2 + \mathsf{Var}(X) = 5 $$

Answer 2

Do you mean $\le$ or $\ge$ in second formula? If $\le$ you can use the Chebyshev inequality:

$$P(|X-\mu|\ge k\sigma) \le 1/k^2.$$

Then substituting your values:

$$P(|X-2|\ge k) \le 1/k^2.$$

If $X$ is positive valued, $P(|X-2|\ge k) = P(X - 2 \ge k)$ then:

$$P(X \ge k+2) \le 1/k^2$$

So, substituting k=3, we have:

$$P(X \ge 5) \le 1/9.$$