Random process involving CDF and PDF of standard normal

1. Jan 7, 2016

JohanL

1. The problem statement, all variables and given/known data
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}

2. Relevant equations

3. solution
A 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.

2. Jan 7, 2016

andrewkirk

Have you learned Wiener processes yet? If you have then a description of a solution that is (to me) much simpler than the above is
$$X(t)=\frac{W(t)}{\sqrt{t}}$$
where $W(t)$ is a Wiener process (aka Brownian Motion).

By the way, I'm pretty sure that line 2 in the problem statement is wrong. It should be:

$$\int_{-\infty}^{0} P\left(X(2)-\frac{1} {\sqrt{2}}X(1)≤-\frac{1} {\sqrt{2}}x\ \bigg\vert\ X(1)=x\right)\phi(x) dx\ \ \ \textbf{[2a]}$$

otherwise the step from line (1) to (2) is not valid.
I think what they expect you to do to validate the red part, is to evaluate$E\left[X(1)\left( X(2)− (\frac{1}{\sqrt 2}) X(1) \right)\right]$. You should be able to show this is zero. However, it would be a mistake to think that means they are independent. Zero covariance is a necessary but not sufficient condition for independence.

So I suggest you forget their solution, which looks dodgy to me. Instead, consider the following:

The step from (1) to (2a) is true if $X(1)\sim N(0,1)$. [Condition A]
The step from 2a to 3 is valid if $X(2)− (\frac{1}{\sqrt 2}) X(1)\sim N(0,1)$ and is independent of $X(1)$. [Condition B]
All other steps are identities, and hold regardless of the distribution.

For a Wiener Process $W(t)$, we have $W(t)\sim N(0,t)$ and $W(t)-W(s)$ is independent of $W(s)$ for $t>s$. So conditions A and B can easily be shown to be met for the process $X(t) =\frac{W(t)}{\sqrt{t}}$. Hence $X(t)$ is a solution.

3. Jan 7, 2016

Ray Vickson

Letting $X_1 = X(1)$ and $X_2 = X(2)$, the random variable $Y = X_2 - a X_1$ has mean 0 and variance
$$\text{Var}(Y) = \text{Var}(X_1) + a^2 \, \text{Var}(X_1) - 2 a \, \text{Cov}(X_1, X_2)$$
What do you get when $\text{Var}(X_1) = \text{Var}(X_2) = 1, a = 1/\sqrt{2}$ and $\text{Cov}(X_1 X_2) =1/ \sqrt{2}\;$?

So, you want $X_2 - X_1/\sqrt{2}$ to be independent of $X_1$ and to have variance 1/2; that is, you want
$$X_2 = X_1/\sqrt{2} \pm Z/\sqrt{2},$$
where $Z \sim \text{N}(0,1)$ is independent of $X_1$. Can you do this in terms of $W(1)$ and $W(2)\:$? What sign in $\pm$ must you take in order to get the result in eq.(3)?