Suppose that two electronic components in the guidance system for a missile operate indepen- dently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.
My solution:
Since $\lambda = 1$, $f(y_1) = e^{-y_1}$ and $f(y_2) = e^{-y_2}$ for $y_1 > 0, y_2 > 0$. Since they operate independently, we have
$$ f(y_1,y_2) = f(y_1)f(y_2) = e^{-y_1} e^{-y_2} $$
The random variable of interest is $U = \frac{Y_1 + Y_2}{2}$. As such, we seek the distribution function $F_U(u) = f_U(U < u) = f_U(\frac{Y_1 + Y_2}{2} < u)$
$$ = \iint_{\frac{Y_1 + Y_2}{2}<u}f(y_1,y_2)dy_1dy_2 $$
$$ = \int_0^{2u} \int_0^{2u-y_1}e^{-y_1} e^{-y_2} dy_2dy_1 = -e^{-2u} -2ue^{-2u} +1 $$
Therefore, we have that
$$ F_U(u) = \begin{cases} 0 & u < 0\\ -e^{-2u} -2ue^{-2u} +1 & u \geq 0\\ \end{cases} $$
therefore, for $u \geq 0$, $$ f_U(u) = \frac{dF_U(u)}{du} = 2e^{-2u} + 4ue^{-2u} $$
The answer, however, is $4ue^{-2u}$. I don't understand where the extra $2e^{-2u}$ came from. Where did I go wrong?
I think you computed the derivative incorrectly. $$\frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$