A continuous random variable X has probability density function
$$f(x)=\begin{cases}
\dfrac{2}{x^{3}}&,x\geq1;\\0&,\text{otherwise}.\end{cases}$$
Find the value of $p\ (0<p<1)$ for which $E(X^p)=\frac{3}{2}$.
I tried to integrate $x^p \frac{2}{x^3}$, but I got $p= -2/3$, which must be wrong. Can anyone please tell me the way of finding $p$?
Best Answer
You probably forgot to carry a negative sign or something.
We have $$ \int_1^\infty x^p \frac{2}{x^3} \, \mathrm{d}x = - \frac{2}{p-2}, p<2 $$
Setting the result equal to $\frac{3}{2}$ yields $p=\frac{2}{3}$.