correlation of product with its normally distributed factors 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?
 A: Given: Random variables $X$ ~ $N(\mu_1,\sigma_1^2)$ and $Y$ ~ $N(\mu_2,\sigma_2^2)$ are independent with joint pdf $f(x,y)$:

(source: tri.org.au) 
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:

(source: tri.org.au) 
Given some numerical values, for instance:

(source: tri.org.au) 
... 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.
