Let $X$ be a an exponentially distributed random variable with rate $\lambda $ and let $t$ be a constant.

It was shown to me that $$E(X-t \; |\; X>t) = E(X)=\frac{1}{\lambda }$$

However, the steps were not shown how to get from the first to the second part. I understand the memoryless property of the exponential distribution but I do not quite see how they are utilizing this to go from the first to the second part.

I am hoping someone can clarify the intuition behind this.

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    $\begingroup$ stats.stackexchange.com/questions/48496/… $\endgroup$ – George Dewhirst Apr 1 at 15:47
  • $\begingroup$ @GeorgeDewhirst I see now. It is the shift and then the reverse shift that cancel. Thank you for the link $\endgroup$ – Jac Frall Apr 1 at 15:50
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    $\begingroup$ $$ E(X-t \; |\; X>t) = E(X \; |\; X>t) - E(t \; |\; X>t) =E(X) + t -t = E(X) =\frac{1}{\lambda } $$ where we use linearity of expectation and then the memoryless property of the Exponential. $\endgroup$ – HJ_beginner Apr 1 at 16:29
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    $\begingroup$ @HJ_beginner your comment is very helpful. It is exactly what my thought was. $\endgroup$ – Jac Frall Apr 2 at 13:29

The (originally) accepted answer is wrong -it has two errors than cancel out- , it could be fixed but it would still be too complex. The comment of HJ_beginner fits the bill.

First: $$E(X-t \; |\; X>t) = E(X \; |\; X>t) - E(t \; |\; X>t) $$

This, because of linearity of (conditional or not) expectation: $E(A+B \mid C)= E(A \mid C) + E(B\mid C)$

Then, $E(t \; |\; X>t)=t$ (expectation of a constant).

And the memoryless property implies that the statistics of $X$ conditioned on $X>t$ are equivalent to the original but shifted a "time" $t$. Hence $E(X \; |\; X>t) = E(X) + t = 1/\lambda +t$.

Hence $E(X-t \; |\; X>t) = E(X)=1/\lambda$



Set $Y=X\cdot \boldsymbol 1_{\{X>a\}}$. Then $$\mathbb E[X\mid X>a]=\frac{1}{\mathbb P\{X>a\}}\mathbb E[X\cdot \boldsymbol 1_{\{X>a\}}]=\frac{1}{\mathbb P\{X>a\}}\mathbb E[Y]=\frac{1}{\mathbb P\{X>a\}}\int_{\mathbb R}yf_Y(y)dy.$$

Now, $$\mathbb P\{Y>y\}=\mathbb P\{X\cdot \boldsymbol 1_{\{X>a\}}>y\}=\mathbb P\{X>y\mid X>a\}\mathbb P\{X>a\}=\mathbb P\{X>(y-a)+a\mid X>a\}\mathbb P\{X>a\}=\mathbb P\{X>y-a\}\mathbb P\{X>a\},$$ where the last equality comes from the memoryless property. Therefore, $$f_Y(y)=\mathbb P\{X>a\}f_{X}(y-a).$$

Now, it's just calculation.

  • $\begingroup$ You have to calculate $\Bbb E[X-a\ \vert\ X>a].$ Right? Though that can be calculated using linearity of conditional expectation. $\endgroup$ – Dbchatto67 Apr 1 at 16:37
  • $\begingroup$ The fact that $\mathbb E[X\mid X>a]=\mathbb E[X]+a$ was not clear for me as far as I did the calculations. But indeed, if it's clear for you, just do it :-) @Dbchatto67 $\endgroup$ – user657324 Apr 1 at 16:42
  • $\begingroup$ It is due to the memoryless property of the exponential distribution @user657324 which is a very standard result. Anyway your answer is good enough to get an upvote at least which no one has given yet. So I upvoted it. $\endgroup$ – Dbchatto67 Apr 1 at 16:54
  • $\begingroup$ are you sure? I think $$\mathbb E[X\mid X>a]=\frac{\mathbb E[X\cdot \boldsymbol 1_{\{X>a\}}]}{P(X>a)}$$ since $$E(X|A)=\frac{E(X1_A)}{P(A)}$$ $\endgroup$ – masoud Apr 2 at 1:59
  • $\begingroup$ but you write $$\mathbb E[X\mid X>a]=\mathbb E[X\cdot \boldsymbol 1_{\{X>a\}}]$$ $\endgroup$ – masoud Apr 2 at 2:36

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