# August 2015, Part 2

Part 1 , 2

## Solution 1:

a) $\lambda_{x}$ (as Y is is Poisson r.v).

(b) For Poisson r.v., $E[Y_{x}]=Var[Y_{x}]=\lambda_{x}$

(c) The attenuation of photons obeys:

$\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}$

(d) The solution to c) gives $\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}$

(e) Given (d), and substitute x with t we can get that $\int_0^x \mu(T)\partial t=-log{\frac{\lambda_{T}}{\lambda_{0}}}=log{\frac{\lambda_{0}}{\lambda_{T}}}$

## Solution 2:

a) Since $Y_{x}$ is Poisson random variable, $E[Y_{x}]=\lambda_{x}$.

(b) For Poisson r.v., $E[Y_{x}]=Var[Y_{x}]\\ \Rightarrow Var[Y_{x}]=\lambda_{x}$

(c) The attenuation of photons obeys:

$\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}$

(d) The solution is:

$\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}$

(e) Based on the result of (d)

$\lambda_{T}=\lambda_{0}e^{-\int_0^T \mu(t)\partial t}\\ \Rightarrow \frac{\lambda_{T}}{\lambda_{0}}=e^{-\int_0^T \mu(t)\partial t}\\ \Rightarrow \int_0^T \mu(t)\partial t=-ln{\frac{\lambda_{T}}{\lambda_{0}}}=ln{\frac{\lambda_{0}}{\lambda_{T}}}$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood