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Let $V_1$,$V_2$, $r$ be independent random variables, where $V_1$, $V_2$ are Gausian with the same distribution and $r$ is uniformly distributed in $[0,1]$ if

$$X(t)=V_2 I(r \geqslant t)+V_1 I(r<t)$$

a)find the mean of this process:

For the expectation I have the following:

$\mathbb{E}X(t)=μ(1-t)+μ t=μ$

b) find the autocorrelation function of this process

$R(t,s)=\mathbb{E}( X(t) X(s) )$

I have simplified this as much as I can, I don't know why I can't copy and paste what I have done, when I try only some of it appears. I have simplified it down to $\sigma^2$ * probability of union of the indicator functions and a similar expression with the mean.

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The autocorrelation function R of the process X is not defined by R(t,s)=E(X(t)X(s)). –  Did May 20 '12 at 8:31
    
@Didier The definition of the autocorrelation function adopted in signal processing is as stated by OP (see wiki), and is different from conventional definition adopted in statistics. –  Sasha May 20 '12 at 13:08
    
@Sasha: Thanks, nice to know. –  Did May 20 '12 at 13:15
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1 Answer 1

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Let $\mu$ denote mean of Gaussians $\mu = \mathbb{E}(V_1)=\mathbb{E}(V_2)$. $$ \begin{eqnarray} \mathbb{E}\left( X(t) \right) &=& \mathbb{E}\left( V_2 I(R \geqslant t) + V_1 I(R<t) \right) \stackrel{\text{independence}}{=} \mathbb{E}(V_2) \mathbb{P}( R \geqslant t) + \mathbb{E}(V_1) \mathbb{P}( R < t) \\ &=& \mu \left( \underbrace{\mathbb{P}( R \geqslant t) + \mathbb{P}( R < t)}_{\begin{array}{c}\text{sum of probabilities of}\\ \text{ complementary events}\end{array}}\right) = \mu \end{eqnarray} $$ As to the autocorrelation function: $$\begin{eqnarray} \mathbb{E}\left( X(t) X(s)\right) &=& \mathbb{E}(\mathbb{V_2^2}) \mathbb{E}\left(I(R \geqslant t)I(R \geqslant s)\right) \\ &+& \mathbb{E}(\mathbb{V_1 V_2}) \left( \mathbb{E}\left(I(R \geqslant t)I(R < s)\right) + \mathbb{E}\left(I(R < t)I(r \geqslant s)\right) \right) \\ &+& \mathbb{E}(V_1^2) \mathbb{E}(I(R>t)I(R>s)) \end{eqnarray} $$ Now expectations of product of indicator functions is done by converting them to probabilities and doing some logic: $$\mathbb{E}\left(I(R \geqslant t)I(R \geqslant s)\right) = \mathbb{P}(R \geqslant t, R\geqslant s) = \mathbb{P}(R \geqslant \max(t,s))$$ $$\mathbb{E}\left(I(R\geqslant t) I(R<s)\right) = \mathbb{P}(s > R\geqslant t)$$ $$\mathbb{E}\left(I(r\geqslant s) I(R<t)\right) = \mathbb{P}(t > R\geqslant s)$$ $$ \mathbb{E}\left( I(R < t)I( R<s)\right) = \mathbb{P}(R < \min(t,s)) $$ Hence $$\begin{eqnarray} R(t,s) &=& \mathbb{E}(V^2) \left( \mathbb{P}\left(R \geqslant \max(t,s)\right) + \mathbb{P}\left(R < \min(t,s)\right) \right) + \mu^2 \mathbb{P} \left( \max(t,s) > R \geqslant \min(t,s) \right) \\ &=& \mathbb{E}(V^2) + \left(\mu^2 - \mathbb{E}(V^2)\right) \mathbb{P} \left( \max(t,s) > R \geqslant \min(t,s) \right) \end{eqnarray} $$

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i solved the question but the solution my professor gave me has confused me, he has for 0≤t≤s≤1, P(r<t,r<s)=P(r<t)=1-t –  Peter Asden Jun 5 '12 at 2:22
    
i solved the question but the solution my professor gave me has confused me,i have solved it using this method and i am working through the method he used ie evaluate the 2 cases, case 1 0≤t≤s≤1 etc, he has P(r<t,r<s)=P(r<t)=1-t,and P(r≥t,r≥s)=P(r≥t)=s but for a uniform random variable it should be the other way round shouldnt it?P(r<t,r<s)=P(r<t)=t and P(r≥t,r≥s)=P(r≥s)=1-s, anyway this has me confused –  Peter Asden Jun 5 '12 at 2:30
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