1
$\begingroup$

If $x(t)$ is a zero mean stationary Gaussian process and if $y(t)=x^2(t)$,then $\{y(t)\}$ is called a square law detector process. Now i want to find autocorrelation function $R_{yy}(t_1,t_2)$,that is $$R_{yy}(t_1,t_2)=E({y(t_1)y(t_2)})$$ $$=E(x^2(t_1)x^2(t_2))$$ Now after this step every books give the result like this

$$R_{yy}(t_1,t_2)=E({x^2(t_1)})E({x^2(t_2)})+2E^2({x(t_2)}X (t_2))$$

But i don't know how this derivation came. Can anyone help me here please?

$\endgroup$
  • $\begingroup$ Could you cite a few books where the result is stated? I am not sure that you have transcribed it correctly, $\endgroup$ – Dilip Sarwate Sep 30 '17 at 16:04
  • $\begingroup$ @DilipSarwate this book sir, page no 359 books.google.co.in/books/about/… $\endgroup$ – Rohit Sep 30 '17 at 16:11
  • $\begingroup$ @DilipSarwate here also books.google.co.in/… $\endgroup$ – Rohit Sep 30 '17 at 16:11
  • $\begingroup$ @DilipSarwate Sir can you help with that? $\endgroup$ – Rohit Oct 1 '17 at 1:46
1
$\begingroup$

Let $\sigma^2$ denote the common variance of the random variables comprising the zero-mean stationary Gaussian process $\{X(t)\}$. We want to find
$$R_{Y}(t,s) = E[Y(t)Y(s)] = E\big[(X(t))^2(X(s))^2\big].$$ Now, given that $X(t)=x$, \begin{align} E\big[(X(t))^2(X(s))^2\mid X(t)=x\big] &= E\big[x^2(X(s))^2\mid X(t)=x\big]\\ &= x^2E\big[(X(s))^2\mid X(t)=x\big] \end{align} where we know (since $X(t)$ and $X(s)$ are jointly Gaussian random variables with covariance $R_X(t-s)$ and correlation coefficient $\rho = \sigma^{-2}R_X(t-s)$) that the conditional distribution of $X(s)$ given that $X(t) = x$ is a Gaussian distribution with mean $\rho x$ and variance $\sigma^2(1-\rho^2)$. It follows that \begin{align} E\big[(X(t))^2(X(s))^2\mid X(t)=x\big] &= x^2E\big[(X(s))^2\mid X(t)=x\big]\\ &= x^2 \left(\sigma^2(1-\rho^2) + \rho^2x^2\right).\end{align} Hence, the random variable $E\big[(X(t))^2(X(s))^2\mid X(t)]$ equals $(X(t))^2 \left( \sigma^2(1-\rho^2) + \rho^2(X(t))^2\right)$ and the law of iterated expectation gives

\begin{align} E\big[(X(t))^2(X(s))^2\big]&=E\bigr[E\big[(X(t))^2(X(s))^2\mid X(t)\big]\bigr]\\ &= E\big[(X(t))^2\left(\sigma^2(1-\rho^2) + \rho^2(X(t))^2\right)\big]\\ &= \sigma^2(1-\rho^2)E[(X(t))^2] + \rho^2 E[(X(t))^4]\\ &= \sigma^4\left((1 - \rho^2) + 3\rho^2\right)\\ &= \sigma^4\left(1 + 2\rho^2\right)\\ &= (\sigma^2)^2 + 2 (\rho\sigma^2)^2\\ &= \left(R_X(0)\right)^2 + 2 \left(R_X(t-s)\right)^2\\ &= E[(X(t))^2]E[(X(s))^2] + 2\left(E[X(t)X(s)]\right)^2. \end{align} This differs in subtle ways (not just the use of $t$ and $s$ in place of $t_1$ and $t_2$) from what the OP insists is the right answer as stated in several books that he claims to have read, which answer he has faithfully transcribed into his question above.

$\endgroup$
0
$\begingroup$

$Cov(y(t1),y(t2)) = E(y(t1)y(t2)) - E(y(t1))E(y(t2))$.

So, this means that

$ Cov(x^2(t1)x^2(t2)) = E(x^2(t1)x^2(t2)) - E(x1^2(t1))E(x2^2(t2))$

Now move the last term on the RHS over to the LHS which gives:

$E(x^2(t1)x^2(t2)) = Cov(x^2(t1)x^2(t2)) + E(x^2(t1))E(x^2(t2))$.

So it comes down to sowing that $Cov(x^2(t1)x^2(t2)) = $ something. But the something can't be what you have in your question. I think you have a typo in the last term of the expression for $R_{yy}(t_{1},t_{2})$. Even then, I'm not sure if it can be easily shown that they ( the something and Cov) are equal but atleast you'll have the right starting point. Good luck.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.