Conditional Mean and Variance of an Exponential Random Variable

The phrase “conditional mean and variance of an exponential random variable” needs a conditioning event to give a unique answer. A common actuarial and reliability question is the expected value and variance of a waiting time or claim amount given that it exceeds a threshold. For that case, the exponential distribution’s memoryless property gives a simple result.

Let $X\sim\operatorname{Exp}(\lambda)$, where $\lambda>0$ is the rate. If the condition is $X>c$ for $c\ge 0$, then

The mean includes the threshold already exceeded, but the variance of the remaining excess is unchanged.

Why the exceedance formulas hold

An exponential random variable with rate $\lambda$ has density and survival probability

For $y\ge 0$, the conditional survival probability of the excess is

This is the survival function of an exponential random variable with rate $\lambda$. Thus, conditional on exceeding $c$, $X$ has the same distribution as $c+Z$, where $Z\sim\operatorname{Exp}(\lambda)$. Since $\mathbb{E}[Z]=1/\lambda$ and $\operatorname{Var}(Z)=1/\lambda^2$, adding the constant $c$ increases the mean by $c$ and does not change the variance.

This property is called memorylessness: after waiting until time $c$, the additional waiting time has the original exponential distribution.

Worked example

Suppose a waiting time has rate $\lambda=0.2$ per hour, so its unconditional mean is $1/0.2=5$ hours. Find its conditional mean and variance given that the wait lasts more than 10 hours.

Set $c=10$ and substitute into the exceedance formulas:

The conditional standard deviation is $\sqrt{25}=5$ hours. In this example, the expected total wait given that it has passed 10 hours is 15 hours; the expected additional wait is 5 hours. The probability of reaching the 10-hour threshold is $\Pr(X>10)=e^{-0.2(10)}=e^{-2}$, about 0.1353. That probability affects how often the condition occurs, but not the conditional excess distribution.

If the condition is a bounded interval

A different answer is needed if you mean conditioning on $c<X<d$, where $0\le c<d$. This is a truncated exponential distribution. Let $L=d-c$; given the interval, the shifted value $Y=X-c$ lies between 0 and $L$. Its conditional mean and variance are

For example, with rate $0.2$ and condition $10<X<15$, $L=5$. The formulas give a conditional mean of approximately $12.09$ hours and variance of approximately $1.98$ hours squared. These differ from the exceedance results because values above 15 hours are excluded.

Taking $c=0$ in the interval formulas gives the case $X<d$. For a very wide interval, as $d$ tends to infinity, the formulas approach the exceedance results with threshold $c$.

Common mistakes

• Confusing rate and scale: Here $\lambda$ is the rate, so the unconditional mean is $1/\lambda$. If a problem uses scale $\theta=1/\lambda$, the exceedance formulas become $c+\theta$ for the mean and $\theta^2$ for the variance.
• Reporting the excess instead of the total: $\mathbb{E}[X-c\mid X>c]=1/\lambda$, while $\mathbb{E}[X\mid X>c]=c+1/\lambda$.
• Assuming every condition preserves memorylessness: The result applies to conditioning on exceeding a threshold. Conditioning on an interval, on being below a threshold, or on another event generally changes the distribution and requires a separate calculation.
• Changing the variance when shifting: Adding the fixed threshold changes the mean, not the variance.