| Line 12: | Line 12: | ||
== Solution 1: == | == Solution 1: == | ||
| − | a) <math> \lambda_{x}</math>. | + | a) <math> \lambda_{x}</math> (as Y is is Poisson r.v). |
(b) For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]=\lambda_{x} | (b) For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]=\lambda_{x} | ||
| Line 21: | Line 21: | ||
<math>\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}</math> | <math>\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}</math> | ||
| − | (d) | + | (d) The solution to c) gives |
<math>\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}</math> | <math>\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}</math> | ||
Revision as of 23:55, 3 December 2015
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
August 2015, Part 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) 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}}} $
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}}} $
