Here's the problem statement :
“Let $X_1$ and $X_2$ be two exponentially distributed random variables with mean $0.5$ and $0.25$. What is the mean of $Y = \min(X_1, X_2)$? ''
We know, the probability density function of exponential function is as follows :
$f(x) = \lambda e^{-\lambda x}$ ; $x \geq 0$ and $0$ everywhere else.
Also, $E(x) = \frac{1}{\lambda}$ and hence $\lambda_1 = \frac{1}{0.5} = 2$ and $\lambda_2 = \frac{1}{0.25} = 4$
The official solution is as follows :
$$E(Y) = \frac{1}{\lambda_1 + \lambda_2} = \frac{1}{2 + 4} = \frac {1}{6}$$
Please help me understand the solution !
Thank you so much.
Best Answer
$$E[Y] = \int_0^{\infty}y\,f_Y(y)\,dy$$
$$f_Y(y)=\frac{dF_Y(y)}{dy}$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=\text{P}\left[\min(X_1, X_2)\leq y\right] \\= 1- \text{P}\left[\min(X_1, X_2) > y\right] \\= 1-\text{P}\left[X_1 > y\right]\text{P}\left[X_2 > y\right]\\=1-\left[1-\text{P}\left[X_1\leq y\right]\right]\left[1-\text{P}\left[X_2\leq y\right]\right]\\=1-\exp\left(-y[\lambda_1+\lambda_2]\right)$$
So
$$f_Y(y) = (\lambda_1+\lambda_2)\exp\left(-y[\lambda_1+\lambda_2]\right)$$
Substitute in the first integration for the expectation and do the integration.