probability density of max(const, X+Y) of random variable X and Y?

Question

Let's say, X and Y are discrete/continuous random variables from gaussian distribution for simplicity. To get the probability density function of f(X,Y), you need to calculate "inverse" of f-function as shown below.

http://www.math.uah.edu/stat/dist/Transformations.html

For continuous case, you need to calculate Jacobian. I am wondering what if function f is like max operator such as f(x)=max(const,X+Y)? I suppose inverse of max(const,X+Y) operator is not even one-to-one function, since output 0 could have been from input of X+Y=-1 or X+Y=-4 or whatever..

Thank you

Answer 1

Step 1. Given the (unstated) joint distribution (see Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?) of $X$ and $Y$, compute the distribution of $Z=X+Y$.

Hint: if $(X,Y)$ are bivariate Gaussian then $X+Y$ is also Gaussian* (though it's not hard to do the integration); given the Gaussian is determined by moments to second order, this is simple enough from basic properties of expectation, variance and covariance.

If you're not able to rely on those, you can compute the required integral readily enough, but if you want to do a bivariate transformation for a bivariate Gaussian, try $(W,Z)=(X-Y,X+Y)$ ... the two will be independent, so you can avoid an integral and the Jacobian is simple enough.

Step 2. Let $M = \max(c,Z)$. Then either the cdf or the pdf for $M$ is straightforward and can essentially be done from first principles. (See wikipedia on truncated distributions)

The approach for other distributions is similar (but in general will offer fewer opportunities for shortcuts in working out the distribution of the sum)

*(but if not bivariate Gaussian, then the sum of the random variables is generally not Gaussian)

(A number of questions on site discuss calculations for step 1 or step 2.)