how to find an actual number for E(X) with the given information?

Question

Let $X$ be a positive continuous random variable with the probability density function $f_X(t)$. Suppose that there is a random variable $Y$, for which the pdf is $f_Y(t) = t\,f_X(t)$ (for all real numbers $t$). What is $E(X)$ (find an actual number)? Express $\operatorname{var}(X)$ in terms of $E(Y)$.

Answer 1

Since $f_Y(t)$ is a pdf, $\displaystyle \int_{-\infty}^\infty f_Y(t)\ dt = 1$. Since $f_Y(t) = tf_X(t)$, then $\displaystyle E(X) = \int_{-\infty}^\infty tf_X(t)\ dt = \int_{-\infty}^\infty f_Y(t)\ dt = 1$.

For the second part, use the fact that $\displaystyle E(Y) = \int_{-\infty}^\infty t^2f_X(t)\ dt = E(X^2)$ to help you write the variance of $X$ in terms of $E(Y)$.