I am wondering if there is a closed form for finding the expected value or variance for a conditional exponential distribution.

For example: $$ E(X|x > a) $$ where X is exponential with mean $\lambda$.

Same question for variance.

What about for a joint distribution of independent exponentials?

$$ E(X|y > a) $$ where X is exponential with mean $\lambda$, Y is exponential with mean $\theta$ and X & Y are independent.

A sample problem for the actuarial P/1 exam (#124 for those also studying) asks:

The joint probability for $f(x,y) = 2e^{-x-2y}, ~ x > 0, ~ y > 0$. Calculate the variance of Y given $x > 3, ~ y > 3$.

The solution goes like this: (Math on the right, reasoning on the left)

  1. $Var (Y|x>3, y>3) =$
  2. $Var (Y|x>3) = ~~~~~$Independence
  3. $Var (Y + 3) = ~~~~~$Memoryless
  4. $Var (Y) + Var (3) =~~~~~$Independence of Y and 3.
  5. $Var (Y) = ~~~~~ $ Since $Var (3) = 0$.
  6. $0.25 ~~~~~ $Exponential Variance, $\lambda = 2$.

So this says to me that $Var (Y|x>3) = Var (Y)$. Is that true? If so, is it always true? If not, then how does this solution work?

Could one also replace E(Y) for Steps 1 - 4, Use $E(a) = a$ and get $E(Y| y>a) = E(y) + a$?

Shortcuts like this are immensely valuable for a timed test. (Not just faster, but less error prone).


Let $X$ be an exponentially distributed random variable with parameter $\lambda$. We can think of $X$ as the lifetime of a specific Carbon $14$ atom. Given that $X\gt a$, the additional lifetime $Y$ of the atom is exponentially distributed with parameter $\lambda$. This is a key property of the exponential, usually called memorylessness.

Given that $X\gt a$, $X=a+Y$. It follows that $E[X|X\gt a]=a+E[Y]=a+E[X]$.

The second question of the same kind is much simpler. If $X$ and $Y$ are any independent random variables, then $E[X|Y\gt a]=E[X]$.

Back to the exponential. Since the random variable $X|X\gt a$ is just $a+Y$, where $Y$ is exponential, we have $\text{Var}(X|X\gt a)=\text{Var}(a+Y)=\text{Var}(Y)=\text{Var}(X)$.

Since we know that an exponentially distributed random variable $X$ with parameter $\lambda$ has mean $\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^2}$, we can write down with no trouble the mean and variance of the conditional version.


The memorylessness property of the exponential distribution states that for if $X$ is exponentially distributed then for any function $f$ we have $E(f(X)|X>\alpha) = E(f(X+\alpha))$. Essentially this means that the distribution of an exponential conditioned to be greater than $\alpha$ is exactly the same as the distribution of an exponential with $\alpha$ added to it.

So your method works for both the mean and the variance. It's also useful to know what happens to the mean and variance of a random variable when you add a constant. (You add the constant to the mean, but leave the variance alone.)

In a maths exam you would be required to prove that the distribution of $X$ and $Y$ are independent exponentials. I don't know if that would be true in an actuary's exam.


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