Expectation of minimum exponentialy distrubited variables

Question

Let's consider two independent variables $X \sim \mathrm{Exp}(4), Y \sim \mathrm{Exp}(12)$. I want to find $$E[\min(4X,Y)]$$

I have my approach but I'm not sure if I'm correct with my way of thinking:

First observation is that : $$ \min(X,Y) = \begin{cases} 4X, & \mbox{if } 4X<Y \\ Y, & \mbox{if } 4X \ge Y \end{cases} $$ So I'm not sure if I'm correct with statement that :

$$E[\min(4X,Y)] = \iint_{\{4x < y\}}4x \cdot \rho_x \,\mathrm{d}x \,\mathrm{d}y + \iint_{\{4x \geq y\}} y \cdot \rho_y \,\mathrm{d}x \,\mathrm{d}y$$ $$= \int_0^\infty \int_0^{\frac{y}{4}} 4x \cdot \rho_x \,\mathrm{d}x \,\mathrm{d} y + \int_0^\infty \int_0^{4x} y \cdot \rho_y \,\mathrm{d} y \,\mathrm{d} x , $$

where $\rho_x, \rho_y$ are the probability density functions of $X$ and $Y$ respectively.

Is my justification correct so far ?

Answer 1

Your approach looks reasonable. The integrals you have look right if you evaluate them. But you should be careful with using the proper notation. For example, the first integral $\iint_{\{4x < y\}} 4x \cdot \rho_{x,y} \,\mathrm{d} x \,\mathrm{d} y$ should still be a double integral with respect to the joint pdf.

To offer a neat alternative approach, we can also use the properties of the Exponential distribution.

• $4 X \sim \mathrm{Exp}(1)$ by closure under scaling by positive factor.
• $\min(4X, Y) \sim \mathrm{Exp}(1 + 12)$ by distribution of the minimum of independent Exponential random variables.