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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?

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

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  • $\begingroup$ Why do we need to include P(X< z, Y< z)? $\endgroup$ – jeff123 May 3 '20 at 13:28
  • $\begingroup$ 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: $X,Y \leq z$). $\endgroup$ – SpiritLevel May 4 '20 at 2:36
  • $\begingroup$ 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 $P(X\wedge Y \leq z)$. $\endgroup$ – SpiritLevel May 4 '20 at 5:06
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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)")
 curve(dexp(x, 9), add=T, lwd=2, col="red")

enter image description here

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.

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