# Conditional Expected Value and distribution question

The distribution of loss due to fire damage to a warehouse is:

$$\begin{array}{r|l} \text{Amount of Loss (X)} & \text{Probability}\\ \hline 0 & 0.900 \\ 500 & 0.060 \\ 1,000 & 0.030\\ 10,000 & 0.008 \\ 50,000 & 0.001\\ 100,000 & 0.001 \\ \end{array}$$

Given that a loss is greater than zero, calculate the expected amount of the loss.

My approach is to apply the definition of expected value:

$$E[X \mid X>0]=\sum\limits_{x_i}x_i \cdot p(x_i)=500 \cdot 0.060 + 1,000 \cdot 0.030 + \cdots + 100,000 \cdot 0.001=290$$

I am off by a factor of 10--The answer is 2,900. I am following the definition of expected value, does anyone know why I am off by a factor of $1/10$?

Should I be doing this instead???

$E[X \mid X>0] = \sum\limits_{x_i} (x_i \mid x_i > 0) \cdot \cfrac{\Pr[x_i \cap x_i>0]}{\Pr(x_i > 0)}$ Thanks.

• The conditional probabilities given $X\gt 0$ are (except for the first one) the probabilities, divided by the probability that $X\gt 0$, that is, by $0.1$. – André Nicolas Jun 10 '13 at 2:04
• I'm trying to figure out if most warehouse fires do nothing, or if warehouses are on fire 10% of the time. – Lucas Jun 11 '13 at 5:03

For example the conditional probability that $X=500$ given that $X>0$ is $$\Pr(X=500\mid X>0) = \frac{\Pr(X=500\ \&\ X>0)}{\Pr(X>0)} = \frac{\Pr(X=500}{\Pr(X>0)} = \frac{0.060}{0.1} = 0.6.$$
You have to divide by $\sum p(x_i)$.
• Why? Is this the proper def I should be using: $E[X \mid X>0] = \sum\limits_{x_i} (x_i \mid x_i > 0) \cdot \cfrac{\Pr[x_i \cap x_i>0]}{\Pr(x_i > 0)}$ – user1527227 Jun 10 '13 at 1:37