The density of a random variable $X$ is $f(x)$ proportional to $x^{-1/2}$ , what is the mean of $X$?

Question

The density of a random variable $X$ is

$f(x)$ proportional to $x^{-1/2}$ for $x \in [0,1]$$

and $f(x) = 0$ for $x \notin [0,1]$. Then, the mean of $X$ is

1. $\frac 12$
2. $\frac 1{\sqrt2}$
3. $\frac 13$
4. $\frac 14$
5. None of the above is correct.

By the formula $\int_{0}^1 x\times x^{-1/2} dx $ (the formula of the expectation of continuous r.v.), I calculate the answer is $\frac 23$, but what is the meaning of the words proportional to? If I multiply some number, option 1-4 are both correct, so the answer is Option 5?

Answer 1

Using the fact that the density must integrate to one over $[0,1]$, then proportional means $$1 = \int_0^1cf(x)\,dx = \int_0^1\frac{c}{\sqrt x}\, dx = c\left[2\sqrt x\right]_0^1 = 2c.$$ Thus $c = 1/2$.

If you now try to compute the expectation, you will find $$\int_0^1x\cdot \frac{1/2}{\sqrt x} = \frac{1}{3},$$ which is option 3.