What Are the Properties of the Maximum and Minimum of Exponential Random Variables?

Let $X_1,\ldots,X_n$ be independent exponential random variables. This assumption matters: knowing only their individual exponential distributions does not determine the distribution of their minimum or maximum if they may be dependent.

We use the rate parameterization: $X_i\sim\operatorname{Exp}(\lambda_i)$ means $X_i$ has density $f_i(t)=\lambda_i e^{-\lambda_i t}$ for $t\ge 0$, and mean $1/\lambda_i$. Some texts use a scale parameter instead; the scale is the reciprocal of the rate.

The minimum is exponential

Define $X_{(1)}=\min(X_1,\ldots,X_n)$. For $t\ge 0$, the minimum exceeds $t$ exactly when every variable exceeds $t$. Independence gives

Thus $X_{(1)}$ is exponential with rate $\sum_{i=1}^n\lambda_i$. In particular,

A related result is that the probability that $X_i$ is the smallest is $\lambda_i/(\sum_j\lambda_j)$. For independent continuous variables, ties have probability zero. A variable with a larger rate is more likely to be the first to occur, because a larger rate means a smaller average waiting time.

The maximum’s distribution

Define $X_{(n)}=\max(X_1,\ldots,X_n)$. The maximum is at most $t$ precisely when all variables are at most $t$, so for $t\ge0$,

For $t<0$, this CDF is zero. Differentiating the product gives its density for $t>0$:

Unlike the minimum, the maximum is generally not exponential. For example, if all rates are equal to $\lambda$, its CDF is $(1-e^{-\lambda t})^n$, which is not the CDF of an exponential variable when $n>1$.

For possibly unequal rates, its expectation can also be written using inclusion–exclusion:

This follows by integrating $\Pr(X_{(n)}>t)=1-\prod_i(1-e^{-\lambda_i t})$ from zero to infinity.

Identically distributed variables: mean and variance

When $X_1,\ldots,X_n$ are independent and all have rate $\lambda$, the ordered values are $X_{(1)}\le\cdots\le X_{(n)}$. Their successive spacings are independent exponentials: the spacing from $X_{(k-1)}$ to $X_{(k)}$ has rate $(n-k+1)\lambda$, taking $X_{(0)}=0$. The maximum is the sum of all these spacings. Therefore,

where $H_n=\sum_{j=1}^n 1/j$ and $H_n^{(2)}=\sum_{j=1}^n 1/j^2$. The variance formula uses the independence of the spacings and the fact that an exponential variable with rate $r$ has variance $1/r^2$.

Worked example: three variables with rate 2

Suppose $X_1,X_2,X_3$ are independent with $X_i\sim\operatorname{Exp}(2)$. Then the minimum has rate $2+2+2=6$, so

The maximum has CDF $F(t)=(1-e^{-2t})^3$ for $t\ge0$, and its density is $6e^{-2t}(1-e^{-2t})^2$. Since $H_3=11/6$ and $H_3^{(2)}=1+1/4+1/9=49/36$,

The minimum is much shorter on average than any one variable, while the maximum is longer; neither result says the maximum itself is exponential.

Common mistakes

• Confusing rate and scale. With rate $\lambda$, the mean is $1/\lambda$. If a problem gives the mean as a parameter, convert it to a rate before using these formulas.
• Assuming the maximum is exponential. The product formula gives its CDF; for $n>1$, that CDF is not exponential.
• Forgetting independence. Multiplying survival probabilities or CDFs requires independence. Without it, the joint distribution is needed.
• Using the equal-rate moment formulas for unequal rates. The harmonic-sum expressions apply when every variable has the same rate. Use the general CDF or inclusion–exclusion expression when rates differ.