Is the Memoryless Property of Exponentials Affected by an Upper Bound?

[hellokitten]

Homework Statement

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?

The Attempt at a Solution

 

[haruspex]

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?

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.

 

[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.

 

[HallsofIvy]

"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 $$\frac{\int_a^b e^{-.5t}dt}{\int_2^9 e^{-.5t}dt}$$

 

[haruspex]

The expected value integral looks okay to me.

Yes, the integral is ok in itself, but

... divided by the (cdf at 9 - cdf at 2)

Exactly - the division is missing.

 

[Ray Vickson]

Homework Statement

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?

The Attempt at a Solution

Homework Statement

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?

The Attempt at a Solution

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.