# Conditional expectation of Exp(theta)

Given X follows an exponential distribution $$\theta$$

how could i show something like

$$\operatorname{E}(X|X \geq \tau)=\tau+\frac 1 \theta$$

?

i have get the idea of using Memorylessness property here,
but how can i combine the probabilty with the expectation?

thanks.

casper

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i have get the idea of using Memorylessness property here,
but how can i combine the probabilty with the expectation?
If you write down mathematically the Memorynessless property then it might become obvious. Otherwise it's fine to express the conditional expectation as a ratio of integrals and evaluate it directly.

If you write down mathematically the Memorynessless property then it might become obvious. Otherwise it's fine to express the conditional expectation as a ratio of integrals and evaluate it directly.
$$\Pr(T > s + t\; |\; T > s) = \Pr(T > t) \;\; \hbox{for all}\ s, t \ge 0.$$

i just cant convert from expectation to the probability...

damn

That's what I did so far.

But I just cant use the memoryless property to do it ....

http://img138.imageshack.us/img138/5945/tempz.jpg [Broken]

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That's what I did so far.

But I just cant use the memoryless property to do it ....
Hint: rewrite the memoryless property into a form suitable to use on line 1 of your proof. Conditional probabilities -> conditional cdf -> conditional pdf.