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Let $T$ be an exponential random variable with parameter $\theta$. For $t \gt0$, compute $\Bbb{E}(T|T\le t)$.

My work: First $$P(T\le s|T\le t)=\frac{\int_0^s\theta e^{-\theta x}dx}{\int_0^t\theta e^{-\theta x}dx} =\frac{1-e^{-\theta s}}{1-e^{-\theta t}}, \ \ 0\le s\le t$$

Hence $$\Bbb{E}(T|T\le t)=\int_0^t s\cdot \frac{\theta e^{-\theta s}}{1-e^{-\theta t}} ds$$

I am not sure if I make a right calculation, please give a verification.

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Your computation looks correct.

Alternative way: using the absence of memory property, we have $$ \mathrm E[T; T> t] = \mathrm E[T\mid T> t] \mathrm P(T>t) = (t+ \mathrm E[T])e^{-\theta t} = (t+ \theta^{-1})e^{-\theta t}, $$ whence $$ \mathrm E[T\mid T\le t] = \frac{\mathrm E[T; T\le t]}{\mathrm P(T\le t)} = \frac{\mathrm E[T] - \mathrm E[T; T> t]}{1-e^{-\theta t}} \\ = \frac{\theta^{-1} - (t+ \theta^{-1})e^{-\theta t}}{1-e^{-\theta t}} = \theta^{-1} - \frac{t}{e^{\theta t} - 1}. $$

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  • $\begingroup$ I think this is a better way and truely what the exercise want me learn. $\endgroup$ – Jaqen Chou Mar 26 at 10:18
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    $\begingroup$ I would not use the word "better" here. Frankly speaking, your approach is better since it is universal while mine can be applied exclusively to exponential distribution. But yes, one usually learns nothing by applying universal methods (just how to apply them); for this reason, in my lectures I give many alternative approaches, which sometimes shed a light on some additional properties. $\endgroup$ – zhoraster Mar 26 at 13:30
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What you have computed in the first line is $P(T \leq s|T\leq t)$ not, $P(T = s|T\leq t)$. (The latter is $0$). So when you compute the expectation you have to replace $1-e^{-\theta s}$ by its derivative.

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  • $\begingroup$ I made an edit. Is this expectation go in right way? $\endgroup$ – Jaqen Chou Mar 26 at 9:21
  • $\begingroup$ @JaqenChou Yes, now it is right. $\endgroup$ – Kavi Rama Murthy Mar 26 at 9:32

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