1. ## Exponential random variable

Let T be an exponential random variable with parameter lamba.Compute E[T/T<=t]

2. In case you don't know the general form of computing such an expectation, follow THIS link.

3. Thanks for the link but I was looking for someone to show me the steps and explain how the expectation is affected because of the conditioning.

4. Originally Posted by ulysses123
Thanks for the link but was looking for someone to show me the steps and explain how the expectation is affected because of the conditioning.
p(T|T<t)=p(T)/p(T<t) for T<t and zero otherwise

CB

5. Originally Posted by CaptainBlack
p(T|T<t)=p(T)/p(T<t) for T<t and zero otherwise

CB
This is supposed to help you perfectly. Just multiply the values with corresponding probabilities and integrate over zero to infinity.

6. Thanks for your reply captain black. I thought that P(T/T<=t) will become
P(T) because the fact that the waiting time was <t has no effect on the distribution of the time you will wait due to the memorylessness property, but you have P(T/T<=t)=P(T,T<=t)/P(T<=t) written as P(T)/P(T<t).
I know you are most likely correct, but i would like to understand the logic as i am not getting how you arrived at this.

7. Originally Posted by ulysses123
Thanks for your reply captain black. I thought that P(T/T<=t) will become
P(T) because the fact that the waiting time was <t has no effect on the distribution of the time you will wait due to the memorylessness property, but you have P(T/T<=t)=P(T,T<=t)/P(T<=t) written as P(T)/P(T<t).
I know you are most likely correct, but i would like to understand the logic as i am not getting how you arrived at this.
Put the expression for the density and p(T<t) into the expression and you will see that the given expression for p(T|T<t) has the required memoryless property.