- #1
JohanL
- 158
- 0
Homework Statement
Let
$$ \Phi(x)=\int_{-\infty}^{x} \frac{1} { \sqrt{2\pi} } e^{-y^2 /2} dy $$
and $$ \phi(x)=\Phi^\prime(x)=\frac{1} { \sqrt{2\pi} } e^{-x^2 /2} $$
be the standard normal (zero - mean and unit variance) cummulative probability distribution function and the standard normal probability density function, respectively. Find a random process $$(X(t), \, t \in \mathbb{R}) $$ such that the following calculation is valid:
\begin{align}P\left(X(1) ≤ 0, X(2) ≤ 0\right) &= P\left(X(1) ≤ 0, X(2)-\frac{1} {\sqrt{2}}X(1)≤-\frac{1} {\sqrt{2}}X(1)\right)\\&=\int_{-\infty}^{0} P\left(X(2)-\frac{1} {\sqrt{2}}X(1)≤-\frac{1} {\sqrt{2}}x\right)\phi(x) dx\\&=\int_{-\infty}^{0} \Phi(-x)\phi(x) dx =\int_{0}^{\infty} \Phi(x)\phi(x) dx =\Bigg[\frac{\Phi(x)^2} {2}\Bigg]^{\infty}_{0}=\frac{3} {8}\end{align}
Homework Equations
3. solutionA zero-mean unit-variance Gaussian process with
$$E(X(1)X(2)) = (\frac{1}{\sqrt 2})$$
, because for such a process
X(1) and $$X(2)− (\frac{1}{\sqrt 2}) X(1)$$ are independent
and
$$P(X(2)−(\frac{1}{\sqrt 2}) X(1) ≤ \frac{1}{\sqrt 2} x) = \Phi(−x)$$since $$X(2)− (\frac{1}{\sqrt 2}) X(1)$$ is a Gaussian random variable with variance 1/2.
If anyone could explain the sections (a) red, (b) green and (c) blue i would appreciate it.