Intuition behind $E(X-t \; |\; X>t) = E(X)=\frac{1}{\lambda }$ in exponential distribution

Question

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.

Answer 1

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

Answer 2

Hint

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.