Suppose for a random variable $X$ it is given $P(X \ge a)=1-\frac{1}{4}a^2$, for $0\le a\le 2$. what is the expectation of $X$?
Correct answer: $\frac{4}{3}$
I have difficulty solving the problem with the above information, all I know is the formula for calculating expectation for continuous random variable is $E[X]=\int_a^b x \, f(x) \, \mathrm{d}x.$ so I tried to put that in action but I missed $x$ and $f(x)$, I assumed that those are $a$ and $P(X\ge a)$ but I know I am lost here. I want to really understand the problem and solve it.
Notice that $\mathbb P(X\in [0,2])=1$ and if $a\in[0,2]$ then $\mathbb P(X\lt a)=\frac{a^2}4$, from which we deduce the cumulative distribution function. A density is given by $f(t):=\frac t2\chi_{[0,2]}(t)$, from which we can compute easily the expectation.