# Any transformation of Gaussian random variables admits covariance calculation after transformation?

Sigmoid transform of Gaussian random variables does not have an easy calculation of covariance.

Specifically, I'm looking for a function transform $$f: \mathbb{R}\rightarrow [0, 1]$$ such that it can transform a Gaussian variable to a probability.

I have two Gaussian random variables $$(x_1, x_2)$$ with non-zero correlation. After transformation, I want to calculate the correlation of $$(f(x_1), f(x_2))$$.

UPDATE: I'm fitting a model $$f(Gaussian(\mu, \Sigma))$$ on some data. After I do model fitting and obtain $$\mu$$ and $$\Sigma$$, I need to calculate the covariance of my model, so I need to calculate $$cov(f(x_1), f(x_2))$$. The function $$f$$ satisfies the modeling need as long as it has a shape like the sigmoid function. But covariance calculation requires a special form of $$f$$.

Any hints?

• 1. logit is log(p/(1-p)). That's not going to make sense on a normal; you probably mean the antilogit exp(x)/(1+exp(x)] ... 2. to what end? What's the ultimate aim? Are approximations okay? – Glen_b Oct 8 '18 at 2:53
• Yes, I mean antilogit or sigmoid. The ultimate goal is the calculation of the covariance of $(f(x_1), f(x_2))$. – Liping Liu Oct 8 '18 at 3:58
• You said that in your post already -- but if the transformation $f$ is essentially any arbitrary cdf (also as in your post), what purpose does that serve?? What's the point in having the covariance between arbitrarily transformed variates? – Glen_b Oct 8 '18 at 5:27
• Perhaps defining a copula might solve your problem? Are you specifically looking for a closed form solution for the covariance? As Glen_b pointed out, it doesn't make sense to find the covariance of $(f_1, f_2)$ after applying any arbitrary transformation $f$. Perhaps what you could do is to apply the appropriate marginal CDFs to transform $$(\Phi_1(x_1), \Phi_2(x_2)) \mapsto (u_1, u_2)$$. You'd then be able to calculate some measure of correlation like the pearson or the kendall tau – InfProbSciX Oct 8 '18 at 10:24
• No, the normal cdf doesn't have an analytical expression. This exercise seems a bit convoluted, even slightly different-looking sigmoid-like functions can enforce very different behavior. Even if you do find a function, be careful with how you use it. E.g. a "quickly varying" sigmoid (one with a massive slope when it curves up) would basically make a lot of your variables go to 0 or 1, and the concept of correlation might not even make sense when applied to such a dataset – InfProbSciX Oct 8 '18 at 19:35

$$f(x) = \left\{\begin{array}{ll} \frac{1}{2}\exp(x) & \mbox{if } x \le 0 \\ 1 - \frac{1}{2} \exp(-x) & \mbox{if } x >0 \end{array} \right.$$
One can take the expectation of $$f(x_1) \cdot f(x_2)$$ with respect to a bi-variate Gaussian distribution of $$(x_1, x_2)$$.