# Expotential distribution

1. Nov 11, 2014

### hellokitten

1. The problem statement, all variables and given/known data
Let x be an exponential random variable with lamda = .5

P(x=5|2<x<9} = 0 since exponential is continuous => probability of any single number = 0
Calculate E[x| 2<x<9] = integral from 2 to 9 of (x* .5*exp(-.5*x)) dx ? is this true or is there something wrong because of the memoryless property of exponentials?

3. The attempt at a solution

2. Nov 11, 2014

### haruspex

No, that equation is not quite right. Imagine reducing the given range to a very narrow one. The expected value should then be some value in that range, but your integral will tend to zero. What have you forgotten?

I don't understand why you think it would violate the memorylessness though. The conditional is not saying anything about the outcome of a previous trial; it's giving you information about the outcome of the present trial.

3. Nov 11, 2014

### RUber

P(5|2<x<9) often is referring to the PDF evaluated at 5 divided by the (cdf at 9 - cdf at 2)
The expected value integral looks okay to me.

4. Nov 11, 2014

### HallsofIvy

Staff Emeritus
"Give that 2< x< 9" means that the probability that x is beween 2 and 9 is 1. The "conditional probability" that "$a\le x\le b$' where, of course, $2< a< b< 9$, is $$\frac{\int_a^b e^{-.5t}dt}{\int_2^9 e^{-.5t}dt}$$

5. Nov 11, 2014

### haruspex

Yes, the integral is ok in itself, but
Exactly - the division is missing.

6. Nov 11, 2014

### Ray Vickson

Because of the upper bound $X < 9$ the memoryless property does not apply in full. However, it applies in part. Conditioned on $X > 2$, the (conditional) distribution of $X$ is the same as that of $2+Y$, where $Y \sim \text{Expl}(\lambda = 0.5)$. We can thus write
$$E(X | 2 < X < 9) = 2 + E(Y|Y < 7) = 2 + E(X | X < 7)$$
Whether or not you think this is useful is for you to decide.

Last edited: Nov 12, 2014