# 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 :

$$E(Y) = \frac{1}{\lambda_1 + \lambda_2} = \frac{1}{2 + 4} = \frac {1}{6}$$

Thank you so much.

• @ZacharySelk its the pdf of exponential distribution $f(x) = \lambda e^{-\lambda x} ; x \geq 0$ and $0$ everywhere else – bluespring Jul 29 '18 at 23:35
• The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $\frac12$ hour and the expected time until a bus on the second route is $\frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $\frac16$ hour – Henry Jul 30 '18 at 0:14
• @Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me. – bluespring Jul 30 '18 at 0:29

$$E[Y] = \int_0^{\infty}y\,f_Y(y)\,dy$$

$$f_Y(y)=\frac{dF_Y(y)}{dy}$$

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.

• @bluespring $$1-\left[1-\text{P}\left[X_1\leq y\right]\right]\left[1-\text{P}\left[X_2\leq y\right]\right] = 1-\left[1-\left\{1-\exp(-\lambda_1\,y)\right\}\right]\left[1-\left\{1-\exp(-\lambda_2\,y)\right\}\right]=1-\exp\left(-y[\lambda_1+\lambda_2]\right)$$ – BlackMath Jul 30 '18 at 2:04
• @bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more. – BlackMath Jul 30 '18 at 3:55

$f(X_2)<f(X_1)$ 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.

• @ZacharySelk Thanks for the correction. – Kavi Rama Murthy Jul 29 '18 at 23:39
• Thank you sir for your kind help. I realize my error now. – bluespring Jul 30 '18 at 23:39

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 $\mathsf P(X_1\leqslant X_2)=\int_0^\infty\int_0^t 8 e^{- 2s-4t}\mathsf d s\mathsf d t=\tfrac 13$

• This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this. – bluespring Jul 29 '18 at 23:53
• but thank you really for your response, its very much appreciated ! – bluespring Jul 30 '18 at 0:00