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If x and y are normally dist. with standard deviation of 10%, and they are independent, then their product X.Y is 71% correlated with Y (or X).

I can show this empirically, but how to I prove it in closed form?

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While your $X$ and $Y$ may be such that $\rho_{XY,X} \approx 0.71$ (shorthand for $1/\sqrt{2} \approx .707\ldots$??), the result is not true in general and so cannot be proved. In particular, if $Y$ has mean $0$, then $\rho_{XY,X} = 0$. On the other hand, it is true that for uncorrelated random variables with equal variance, $\rho_{X-Y,X} = 1/\sqrt{2}$. See this answer on stats.SE for a proof. –  Dilip Sarwate May 25 '13 at 20:35
    
thanks Dilip, liked your solution! but wolfies gave me exactly what I wanted. You are right though that it is not true in general and as wolfies said, my parameter set determines the answer. It just so happened that my 'random' parameter set and the person who claimed the 71% correl had the same characteristics!! what are the chances :D –  Jay May 26 '13 at 7:39

1 Answer 1

Given: Random variables $X$ ~ $N(\mu_1,\sigma_1^2)$ and $Y$ ~ $N(\mu_2,\sigma_2^2)$ are independent with joint pdf $f(x,y)$:

You seek:

$$correlation(X Y, Y) = \frac{\text{Cov}(X Y,\text{}Y)}{ \sqrt{\text{Var} (X Y)} \sqrt{\text{Var} Y}}$$

The solution, obtained here using mathStatica, is simply:

Given some numerical values, for instance:

... the answer would be 2/3 for those particular parameter values. It is easy to do a quick Monte Carlo check:

  xdata = RandomReal[NormalDistribution[2, 1], 1000000];
  ydata = RandomReal[NormalDistribution[2, 1], 1000000];
  Correlation[xdata * ydata, ydata]

0.666552

Looks fine. Your empirical case of 0.71 would presumably fit the parameter values you have been using ...

More generally ...

One can obtain an exact symbolic solution to the correlation you seek, without assuming a specific parametric form (Normality), and without even needing to assume independence. Here, again derived by mathStatica, is the general solution for any 2 independent random variables (whose first 2 moments exist):

$$\text{correlation}(X Y, Y) = \frac{\text{Var}(Y) \mathbb{E}[X]}{\sqrt{\text{Var}(X) (\mathbb{E}[Y])^2+\text{Var}(Y) \left(\text{Var}(X)+(\mathbb{E}[X])^2\right)}\sqrt{\text{Var}(Y)} }$$

The dependent case is just a bit more messy.

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Your last result gives a correlation of $1/\sqrt{3}$ when the two standard deviations and the two means all have the same value $\sigma$ –  Dilip Sarwate May 25 '13 at 20:43
    
thanks guys. however, wolfies did you mean to add that last 1/surd(varY) term? It doesn't seem to match your output from mathStatica above, or is that term on top miu1*sigma2 instead of miu1*(sigma2)^2 because that pic was a little cropped at the top :) –  Jay May 26 '13 at 7:36
    
@DilipSarwate Yes - both the general result and the Normal result (as one would hope!) above are equal to $\frac{1}{\sqrt{3}}$ if the means and standard deviations are all equal. –  wolfies May 26 '13 at 7:39
    
@user79429 Greetings. Yes - everything is correct ... the two solutions match perfectly. To see this, simply substitute in $$\left\{\text{Var}(X)\to \sigma _1^2, \text{Var}(Y)\to \sigma _2^2, \mathbb{E}[X]\to \mu _1, \mathbb{E}[Y]\to \mu _2\right\}$$ to the general case, and you will obtain the Normal case above. –  wolfies May 26 '13 at 8:16

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