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Random process involving CDF and PDF of standard normal

  1. Jan 7, 2016 #1
    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. jcsd
  3. Jan 7, 2016 #2

    andrewkirk

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    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.
     
  4. Jan 7, 2016 #3

    Ray Vickson

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    Letting ##X_1 = X(1)## and ##X_2 = X(2)##, the random variable ##Y = X_2 - a X_1 ## has mean 0 and variance
    [tex] \text{Var}(Y) = \text{Var}(X_1) + a^2 \, \text{Var}(X_1) - 2 a \, \text{Cov}(X_1, X_2) [/tex]
    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
    [tex] X_2 = X_1/\sqrt{2} \pm Z/\sqrt{2}, [/tex]
    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)?
     
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