Let $Y_1 =\frac{1}{2}X^2 − 1$ and $Y_2 =\frac{1}{2}X − 1$, where $X$ is a random variable whose mean is positive. Moreover, we know that the mean of $Y_1$ is $2$, and the variance of $Y_2$ is $0.5$. Find the mean and variance of $X$.
This seems simple, but I am stumped. Any ideas?
Apply the properties you know of expectation and variance to $$2 =E[Y_1] = E\left[\frac{1}{2}X^2-1\right]=\frac{1}{2}E[X^2]-1$$
and hence $E[X^2]=6$. Next $$\frac{1}{2} = \text{Var}(Y_2) = \text{Var}\left(\frac{1}{2}X-1\right) = \frac{1}{4}\text{Var}(X)$$ and hence $\text{Var}(X) = 2$. Finally use $$\text{Var}(X) = E(X^2)-\{E(X)\}^2$$ to solve for the positive expectation $E[X] = 2$.