Let X be a non constant positive random variable such that E(X)=9. Then which of the following statements is correct Let X be a non constant positive random variable such that E(X)=9. Then which of the following statements is correct
A) $E(\frac{1}{X+1})>0.1$ and $P(X \geq 10) \leq 0.9$
B) $E(\frac{1}{X+1})<0.1$ and $P(X \geq 10) \leq 0.9$
C) $E(\frac{1}{X+1})>0.1$ and $P(X \geq 10) > 0.9$
D) $E(\frac{1}{X+1})<0.1$ and $P(X \geq 10) > 0.9$
I am unable to understand how to proceed. Can someone please help me out
 A: You need two ingredients:

*

*Markov inequality

https://en.wikipedia.org/wiki/Markov%27s_inequality

*

*Jensen inequality

https://en.wikipedia.org/wiki/Jensen%27s_inequality
You just have to be careful at checking in the second case if the function is convex/concave on the domain and apply the signs accordingly. Can you try ?
A: There are two ways to proceed.
The good way is to use properties of random variables. The first part uses Jensen's Inequality applied to the function $f(x) = 1/(x+1)$ from which it follows that
$$E[f(X)] > f(E[X]) = 1/(9 + 1) = 0.1$$
(it's an inequality because $X$ is not constant.) For the second part, you have
$$9 = E[X] = \sum_x P(X=x) x = \sum_{x < 10} P(X=x) x + \sum_{x \ge 10} P(X=x) x \ge P(X \ge 10) \times 10$$
and so $P(X \ge 10) \le 9/10$. Or, in the continuous case, you can do a similar proof with integrals instead of sums (actually the proof with integrals works in all cases if you use measure theory correctly.)
The evil way is to use the fact that the question has only one answer, so you only need an example of an $X$ which makes one of the four alternatives true. The easiest $X$ is probably $X = 1$ with probability $0.5$ and $X = 17$ with probability $0.5$. Then you see immediately that $$E[1/(X+1)] = 0.5 \times 1/(1+1) + \ldots > 1/4 > 0.1$$ and $P(X \ge 10) = 0.5 \le 0.9$, so if there is a valid answer, it must be (A).
