I'm studying the memorylessness property of exponential r.v.'s. Here's how I understand it (story explanation): The probability that starting from $t_0=0$ light bulb will work $5$ hours and then $2$ extra hours or more equals to the probability that it will work $2$ hours or more starting from $t_0=0$. So far so good (I hope I'm right in understanding the memorylessness property).
Now I'm going further and trying to understand how to compute conditional expectation given that the r.v. is greater than some number.
I 've read in multiple sources that ${\bf E}[X|X>x]=x+{\bf E}[X]$. I'm trying to apply the definitions:
$f_{X|A}(x)=\frac{f_{X}(x)}{P(X>x)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}}=\lambda$
However, if I apply the definition of expectation to that new conditional PDF, the integral will not converge.
I have basically 2 questions:
1) what's the verbal explanation of the formula ${\bf E}[X|X>x]=x+{\bf E}[X]$
2) what's wrong with my math reasoning which leads to a divergent integral?
We know we have a memoryless process with exponential distribution. This immediately means we can write $$ f_{X|X>x} = \frac{\lambda \mathrm{e}^{-\lambda t}}{\mathrm{e}^{-\lambda x}} = \lambda \mathrm{e}^{-\lambda (t-x)} = f_X(t-x) \,\forall\,\, t > x $$ so we want to compute the expectation of this $$ \int_x^\infty t\lambda \mathrm{e}^{-\lambda (t-x)} dt =\int_x^\infty t\lambda \mathrm{e}^{-\lambda (t-x)} dt $$ lets change of variables $u = t-x$ (as per the link above) $$ \int_0^\infty (u+x)\lambda\mathrm{e}^{-\lambda u}du = \int_0^\infty u f_X(u)du + x\int_0^\infty f_X(u)du $$ we know that $$ \int_{\Omega}xf_X(x)dx = \mathbb{E}_X(x) $$ and we know that $$ \int_{\Omega}f_X(x)dx = 1. $$