Given two exponential random variables $X_1$ and $X_2$ and their expected values, how do i approach to find the expected value of $Y = \min(X_1, X_2)$ 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.
 A: $$E[Y] = \int_0^{\infty}y\,f_Y(y)\,dy$$
$$f_Y(y)=\frac{dF_Y(y)}{dy}$$
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.
A: $f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $P\{Y>t\}=P\{X_1>t\}P\{X_2>t\}=\int_t^{\infty} (0.5) e^{-(0.5)s} \ ds \int_t^{\infty} (0.25) e^{-(0.25)s} \ ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation. 
A: If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $\mathsf P(X_1\leqslant X_2)=\int_0^\infty\int_0^t 8 e^{- 2s-4t}\mathsf d s\mathsf d t=\tfrac 13$
