Can't work out this conditional expectation

Question

For some positive $\lambda$ and probability measure defined by $\Pr\left((a,b]\right)=e^{-\lambda a}-e^{-\lambda b}$. We are working with $\Omega=(0,\infty)$ and the $\sigma$-algebra is borel of the same. I am trying to work out $\mathrm{E}\left[X\mid\sigma(Y)\right]$ and $\mathrm{E}\left[e^{-\alpha X}\mid\sigma(Y)\right]$ for $X$ and $Y$ defined as:

$X(\omega)=\omega$,

$Y(\omega)=\min\left\{\omega,\kappa\right\}$,

$\kappa$ is some positive constant.

I haven't made any progress. Any hints as to how I should go about this?

Answer 1

• For $y=\kappa$ we have: $\mathbb{E}\left[X\mid Y=\kappa\right]=\mathbb{E}\left[X\mid X\geq\kappa\right]=\kappa+\frac{1}{\lambda}$.

The RHS is based on the fact that under condition $X\geq\kappa$ random variable $X-\kappa$ again has exponential distribution with parameter $\lambda$. That rests on the fact that exponential distributions are memoryless.

• For $y\in\left(0,\kappa\right)$ we have: $\mathbb{E}\left[X\mid Y=y\right]=y$.

The two bullets together allow the conclusion that: $$\mathbb{E}\left[X\mid Y\right]=Y+\frac{1}{\lambda}\mathbf{1}_{Y=\kappa}$$

Try to find $\mathbb E[e^{-\alpha X}\mid Y]$ on a similar way.