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

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 $\frac{1}{5}$?

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 $\mathsf{Var}(X) = 1$, $$\mathsf{E}(X^2) = \mathsf{E}(X)^2 + \mathsf{Var}(X) = 5$$

• Thank you. This more directly answers the proof that my instructor was looking for and now that I see it in front of me it makes perfect sense. The answer from Proved Maroon Z was not wrong though. Thanks to both of you. – Halpneeded Jun 16 '17 at 14:54

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

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

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

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:

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

• I did mean ≤. I will fix it. As for your response, I have gotten the same result using Chebyshev's inequality, and you would THINK that would be acceptable since if $P(X≥5)≤1/9$ by intuition means $P(X>5)≤1/5$ but my instructor would be a real jerk to give us a trick question. I'm not completely convinced this is the proof he's looking for, especially after giving that weird hint. – Halpneeded Jun 16 '17 at 3:27
• To be fair, you did answer the question, and I thank you for it. I will wait and see if anyone else has a response more in line with what the instructor is searching for. – Halpneeded Jun 16 '17 at 3:33
• Also your hint correctly looks so: $P(|X| ≥ a) = P(\phi(X) ≥ \phi(a))≤\frac{E[\phi(X)]}{\phi(a)}$. And I have to note, that Chebyshev inequality is Markov inequality when you substitute $(Y-\mathbf{E}Y)^2$ instead of $X$. Try to read about Chernov inequality, if you want to prove inequality this way. – Proved Maroon Z Jun 16 '17 at 3:51