0
$\begingroup$

Assuming that we have:

The mean of random variable $x$ noted as $\mu_x$, 2nd moment of $x$ noted as $E[x^2]$.

We have also the mean of $(y|x)$ noted as $A(x)+B$, and its 2nd moment $E[y^2|x]$, with constants $A$, $B$, $E[y^2|x]=C$

What we want are the mean and 2nd moment of $y$.

I think they are possible to compute, as \begin{equation} \begin{split} \int p(y)ydy & =\int \int p(y|x) p(x) y \ dxdy \\ & = \int \left(A(x)+B\right)p(x) \ dx \\ & = A\left( \mu_x \right) + B \end{split} \end{equation}

but for the correlation \begin{equation} \begin{split} \int p(y)y^2dy & = \int \int p(y|x)p(x) y^2 \ dxdy \\ \end{split} \end{equation}

I have no idea how to continue the derivation...

Thank you for your attention

  • 0
    Suggest you start by straightening out terminology 'correlation' and 'standard deviation'. – 2017-02-03
  • 0
    @BruceET, sorry for the mistake, I've modified the question, it is neither Correlation nor Covariance, it is 2nd moment. But if we get the 2nd moment, it is equal to have the Cov. The problem here is we only know the moments of (y|x), if we know the function of (y|x) = Ax+B, it would be easy. – 2017-02-06

0 Answers 0