Difference between revisions of "ECE600 QE 2003 January" - Rhea

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7.6 QE 2003 January

Problem 1 (30 points)

i)

Let $ \mathbf{X} $ and $ \mathbf{Y} $ be jointly Gaussian (normal) distributed random variables with mean $ 0 $ , $ E\left[\mathbf{X}^{2}\right]=E\left[\mathbf{Y}^{2}\right]=\sigma^{2} $ and $ E\left[\mathbf{XY}\right]=\rho\sigma^{2} $ with $ \left|\rho\right|<1 $ . Find the joint characteristic function $ E\left[e^{i\left(h_{1}\mathbf{X}+h_{2}\mathbf{Y}\right)}\right] $ .

• We can find the correlation coefficient using the covariance and variances $ r=\frac{Cov\left(\mathbf{X},\mathbf{Y}\right)}{\sigma^{2}}=\frac{E\left[\mathbf{XY}\right]-E\left[\mathbf{X}\right]E\left[\mathbf{Y}\right]}{\sigma^{2}}=\frac{\rho\sigma^{2}-0\cdot0}{\sigma^{2}}=\rho. $

• Now, we can get the joint characteristic function $ \Phi_{\mathbf{X}\mathbf{Y}}\left(\omega_{1},\omega_{2}\right)=e^{i\left(0\cdot\omega_{1}+0\cdot\omega_{2}\right)}e^{-\frac{1}{2}\left(\sigma^{2}\omega_{1}^{2}+2r\sigma^{2}\omega_{1}\omega_{2}+\sigma^{2}\omega_{2}\right)}=e^{-\frac{1}{2}\sigma^{2}\omega^{2}\left(2+2\rho\right)}=e^{-\sigma^{2}\omega^{2}\left(1+\rho\right)}. $

ii)

Let $ \mathbf{X} $ and $ \mathbf{Y} $ be two jointly Gaussian distributed r.v's with identical means and variances but are not necessarily independent. Show that the r.v. $ \mathbf{V}=\mathbf{X}+\mathbf{Y} $ is independeent of the r.v. $ \mathbf{W}=\mathbf{X}-\mathbf{Y} $ . Is the same answer true for $ \mathbf{A}=f\left(\mathbf{V}\right) $ and $ \mathbf{B}=g\left(\mathbf{W}\right) $ where $ f\left(\cdot\right) $ and $ g\left(\cdot\right) $ are suitable functions such that $ E\left[f\left(\mathbf{V}\right)\right]<\infty $ and $ E\left[g\left(\mathbf{W}\right)\right]<\infty $ . Given reasons.

iii)

Let $ \mathbf{X} $ and $ \mathbf{Y} $ be independent $ N\left(m,1\right) $ random variables. Show that the sample mean $ \mathbf{M}=\frac{\mathbf{X}+\mathbf{Y}}{2} $ is independent of the sample variance $ \mathbf{V}=\left(\mathbf{X}-\mathbf{M}\right)^{2}+\left(\mathbf{Y}-\mathbf{M}\right)^{2} $ . Note: $ \mathbf{V} $ is not a Gaussian random variable.

Problem 2 (35 points)

Consider the stochastic process $ \left\{ \mathbf{X}_{n}\right\} $ defined by: $ \mathbf{X}_{n+1}=a\mathbf{X}_{n}+b\mathbf{W}_{n} where \mathbf{X}_{0}\sim N\left(0,\sigma^{2}\right) $ and $ \left\{ \mathbf{W}_{n}\right\} $ is an i.i.d. $ N\left(0,1\right) $ sequence of r.v's independent of $ \mathbf{X}_{0} $ .

i)

Show that if $ R_{k}=cov\left(\mathbf{X}_{k},\mathbf{X}_{k}\right) $ converges as $ k\rightarrow\infty $ , then $ \left\{ \mathbf{X}_{k}\right\} $ converges to a w.s.s. process.

ii)

Show that if $ \sigma^{2} $ is chosen appropriately and $ \left|a\right|<1 $ , then $ \left\{ \mathbf{X}_{k}\right\} $ will be a stationary process for all $ k $ .

iii)

If $ \left|a\right|>1 $ , show that the variance of the process $ \left\{ \mathbf{X}_{k}\right\} $ diverges but $ \frac{\mathbf{X}_{k}}{\left|a\right|^{k}} $ converges in the mean square.

Problem 3 (35 points)

i)

Catastrophes occur at times $ \mathbf{T}_{1},\mathbf{T}_{2},\cdots $, where $ \mathbf{T}_{i}=\sum_{k=1}^{i}\mathbf{X}_{k} $ where the $ \mathbf{X}_{k} $ 's are independent, identically distributed positive random variables. Let $ \mathbf{N}_{t}=\max\left\{ n:\mathbf{T}_{n}\leq t\right\} $ be the number of catastrophes which have occurred by time $ t $ . Show that if $ E\left[\mathbf{X}_{1}\right]<\infty $ then $ \mathbf{N}_{t}\rightarrow\infty $ almost surely (a.s.) and $ \frac{\mathbf{N}_{t}}{t}\rightarrow\frac{1}{E\left[\mathbf{X}_{1}\right]} $ as $ t\rightarrow\infty $ a.s.

ii)

Let $ \left\{ \mathbf{X}_{t},t\geq0\right\} $ be a stochastic process defined by: $ \mathbf{X}_{t}=\sqrt{2}\cos\left(2\pi\xi t\right) $ where $ \xi $ is a $ N\left(0,1\right) $ random variable. Show that as $ t\rightarrow\infty,\;\left\{ \mathbf{X}_{t}\right\} $ converges to a wide sense stationary process. Find the spectral density of the limit process.

Hint:

Use the fact that the characteristic function of a $ N\left(0,1\right) $ r.v. is given by $ E\left[e^{ih\mathbf{X}}\right]=e^{-\frac{h^{2}}{2}} $ .

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