Markov's inequality results negative

Question

$X$ is the midterm grade (out of $100$ points) of a randomly chosen student. If the average grade is $40$ points, find an upper bound on $Pr[X < 10]$ using Markov's inequality. (Hint: You may need to define a new non-negative random variable.)

I encountered this question and applied Markov's inequality but it gave me $P(X < 10)>= 1-\left(\frac{40}{10}\right)$ and that gives us a negative bound. How can I resolve this problem ?

Answer 1

The problem is that $P( X \leq 10) $ is not of the form desired by the Markov inequality, and using $P(X \leq 10) = 1- P(X \geq 10)$ followed by Markov does not help since the RHS becomes negative, so the inequality is obvious.

To get it to the Markov form, you must get a non-negative random variable $X'$ and a quantity $T > 0$ so that $P(X' \geq T) = P(X \leq 10)$. Then you can apply Markov inequality to $X'$ to conclude.

In this case, for example take $X' = 100-X$, then $E[X'] = 100-E[X]$ and $P(X' \geq 90) = P(X \leq 10)$. Use Markov inequality, conclude.

Answer 2

I think, what you did was: Let $X$ be the grade of a randomly chosen student. We know $0\leq X \leq 100$ and $\mathbb{E}X=40$. We have $$\mathbb{P}(X<10)=1-\mathbb{P}(X\geq10)\geq 1-\frac{\mathbb{E}X}{10}=1-\frac{40}{10}=-3,$$ which is correct, but a completely useless lower bound.

The hint was to define a new nonnegative random variable. So, let us consider $Y:=100-X$, i..e, the points missing for full points. Then, $$\mathbb{P}(X<10)=\mathbb{P}(100-X> 90)= \mathbb{P}(Y\geq 91)\leq\frac{\mathbb{E}Y}{91}=\frac{100-\mathbb{E}X}{91}=\frac{60}{91},$$ which is a more useful upper bound.