# Statistical Hypothesis Testing for $f(x)p(x)+g(y)q(y) \equiv C$

$$f(x)$$, $$p(x)$$, $$g(y)$$ and $$q(y)$$ are four functions, where $$f(x)$$ and $$g(y)$$ are probability density functions. $$C$$ is a constant.

Now I have some statistics $$x_1,...,x_n$$ and $$y_1,...,y_n$$, where $$x_i \sim f(x)$$ and $$y_i \sim g(x)$$. $$p(x_i)$$ and $$q(y_i)$$ are known for all $$i$$.

Is there any statistical hypothesis testing method for the hypothesis $$f(x)p(x)+g(y)q(y) \equiv C$$?

If not, what about a simplified problem where $$p(x) \equiv q(y) \equiv 1$$?

• What is their support? Because your special case doesn't make sense if it is unbounded. – Bartek Jul 11 at 11:07
• How would you interpret $x$ in $f(x)+g(x)$? – Henry Jul 11 at 11:39
• @Bartek their support is [0, 1]. – Zehui Lin Jul 12 at 5:32
• @Henry I edit the problem. $f(x)$ and $g(y)$ now are PDFs for different variables. – Zehui Lin Jul 12 at 5:38

In the special case we have to prove that PDFs od $$x$$s and $$y$$s sum up to a constant. Since their support is $$[0,1]$$ intergating both sides on this interval yields that this constant has to be rwo. If their PDFs sum to one that means their arithmetic mean is one Anne therefore that the arithmetic mean of their CDFs is the function $$f(x)=x$$. So you can just compute ECDFs for both $$x$$s and $$y$$s, add them up, divide by two and compare vs the theoretical CDF via for example Kolmogorov-Smirnov test. In more general case however I don't have an answer.