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?

  • $\begingroup$ 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

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.


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

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.


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