# Distribution of minimum of two independent exponential distributions

Suppose $$\lambda, \mu > 0$$ and $$X \sim \mathsf{Exp}(\lambda)$$ and $$Y \sim \mathsf{Exp}(\mu).$$ We were asked to find the distribution of $$\min(X,Y).$$ Let $$\ell$$ be in R. $$P(\min(X,Y)< \ell)= P(X < \ell , Y > \ell) + P(X>\ell, Y < \ell)$$ But with the above expression, I did not get the correct distribution of $$\min(X,Y).$$ Am I missing out a term from my expression above?

• The term you are missing is $P(X < l, Y < l)$. – Peter May 3 '20 at 5:05

As mentioned in the comments, you are missing one set. Here is how you would find the probability density using your starting point.

Let $$X\sim Exp(\lambda)$$ and $$Y\sim Exp(\mu)$$ be independent. Hence, $$X$$ has pdf $$f_X(x)=\lambda e^{-\lambda x}$$ and cdf $$F_X(x)=1-e^{-\lambda x}$$ on $$[0,\infty)$$. And, $$Y$$ has pdf $$f_Y(y)=\mu e^{-\mu y}$$ and cdf $$F_Y(y)=1-e^{-\mu y}$$.

Let $$Z=X\wedge Y$$ denote the minimum of the two.

$$\begin{eqnarray*} F_Z(z) &=& P(Z\leq z)\\ &=& P(X\wedge Y \leq z)\\ &=& P(X\leq z < Y)+P(Y\leq z < X)+P(X\leq z, Y\leq z)\\ &=& P(X\leq z)P(z < Y)+P(Y\leq z)P(z < X)+P(X\leq z)P(Y\leq z)\\ &=& P(X\leq z)P(z < Y)+P(Y\leq z)\Big(P(z < X)+P(X\leq z)\Big)\\ &=& P(X\leq z)P(z < Y)+P(Y\leq z)\\ &=& (1-e^{-\lambda z})e^{-\mu z}+1-e^{-\mu z}\\ &=& 1-e^{-(\lambda+\mu) z} \end{eqnarray*}$$ and so the pdf is $$f_Z(z)=F'_Z(z)=(\lambda+\mu)e^{-(\lambda+\mu)z}$$.

However, starting with $$P(X\wedge Y>z)=P(X>z,Y>z)=P(X>z)P(Y>z)=e^{-\lambda z}e^{-\mu z}=e^{-(\lambda+\mu)z}$$ is much quicker.

• Why do we need to include P(X< z, Y< z)? – jeff123 May 3 '20 at 13:28
• We need to account for all the different ways $X\wedge Y \leq z$, which are (i) one is below $z$ while the other is above $z$ (there are two ways: $X\leq z<Y$ and $Y\leq z < X$) and (ii) both are below $z$ (there is one way: $X,Y \leq z$). – SpiritLevel May 4 '20 at 2:36
• Here is another way to look at it: Let $\omega\in (X\leq z, Y\leq z)$. This means that $X(\omega)\leq z$ and $Y(\omega)\leq z$ which then means that $X(\omega)\wedge Y(\omega) \leq z$ and hence $\omega\in (X\wedge Y \leq z)$. Thus, $(X\leq z, Y\leq z)$ is a subset of $(X\wedge Y \leq z)$ and so it must be included in any calculation of $P(X\wedge Y \leq z)$. – SpiritLevel May 4 '20 at 5:06

I think it will be easier to start this way: Let $$V = \min(X,Y),$$ where $$X$$ and $$Y$$ are independent exponential random variables with rates $$\lambda$$ and $$\mu,$$ respectively as in the question. Then

$$1-F_V(v) = P(V > v) = P(X > v,\, Y > v)\\ = P(X > v)P(Y > v) = e^{-\mu v}e^{-\lambda v} = \cdots .$$

Then find $$F_v(v)$$ and recognize it as the CDF of another exponential distribution with rate $$\lambda + \mu.$$

Simulation in R where rexp samples from an exponential population, and dexp is an exponential density function. Samples of size $$m = 100\,000$$ are drawn from the distributions of $$X$$ and $$Y.$$ Then the elementwise min is taken of the resulting two $$m$$-vectors to get a random sample from the distribution of $$V.$$

set.seed(2020)  # for reproducibility
x = rexp(10^5, 4);  y = rexp(10^5, 5)
v = pmin(x,y)   # elementwide min
mean(v)
[1] 0.1116067   # aprx E(V) = 1/9 = 1.111

hist(v, prob=T, br=30, col="skyblue2",
main="Simulated Sample from EXP(rate=9)")

The density curve of $$\mathsf{Exp}(\mathrm{rate}= 9)$$ is a good fit to the histogram of the $$m$$ realizations of $$V.$$
Rationale. Suppose components A and B have exponentially distributed lifetimes with rates $$4$$ and $$5$$ failures per year, respectively. Thus their individual expected lifetimes are $$0.25$$ and $$0.20$$ years, respectively.
If A and B are connected in series to make a system that fails when the first (minimum) of A or B fails, then the system has failure rate $$4+5 = 9$$ per year, and an expected lifetime of $$1/9 \approx 0.11$$ years.