Exponential random variable

• Mar 18th 2011, 11:44 PM
ulysses123
Exponential random variable
Let T be an exponential random variable with parameter lamba.Compute E[T/T<=t]
• Mar 18th 2011, 11:51 PM
Sambit
In case you don't know the general form of computing such an expectation, follow THIS link.
• Mar 19th 2011, 01:10 AM
ulysses123
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.
• Mar 19th 2011, 01:21 AM
CaptainBlack
Quote:

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
• Mar 19th 2011, 07:48 AM
Sambit
Quote:

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.
• Mar 19th 2011, 05:20 PM
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.
• Mar 19th 2011, 09:10 PM
CaptainBlack
Quote:

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.