# Problems getting transformation function from source and destination random variables knowledge when handling the discrete case

In this question I asked about a way in order to find a specific transformation function $g(\cdot)$ in order to transform a random variable into another one.

Thanks to the answer to that question I could investigate a little more and understood that $g(\cdot)$ can be found using the CDFs of both distributions. So let's say that I have a random variable $X$ and I want to transform it into $Y$ knowing both $F_Y(y)$ and $F_X(x)$. The transformation $g(\cdot)$ that allows $Y = g(X)$ is the following:

$$g(\cdot) = F_Y^{-1}(F_X(\cdot))$$

### How to handle discrete variables?

But I have a problem. The transformation $F_Y^{-1}(F_X(\cdot))$ cannot be done when I have discrete variables, or better when $Y$ is a discrete random variable.

It is obvious why. Being a discrete random variable $F_Y(y)$ will be a stair function, which is no a bijection! Inversion is not possible!

It seems that it is not possible to transform a generic random variable into a discrete one...

### Question

My questions are:

1. How to handle this, in the case I want $Y$ to be a discrete random variable?

2. Is this method only for those cases where the destination random variable $Y$ is a continuos r.v.? Meaning that the method cannot transform a r.v. into a discrete one.

3. Other than this approach, is there a method to get $g(\cdot)$ in order to achieve $Y=g(X)$ knowing everything about $X$ and $Y$ (moments, PMFs, CDFs and so on...)?

Thankyou

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ex: Convert $Y = ln(X)$.

First of all, we know that $X=e^Y$, so the domain of $Y$ is all real numbers. We also know the pdf of X is $F_X(x) = 1-e^x$. Then

$F_Y(y)=P(Y<y)=P(ln(x)\le Y)=P(x\le e^y)=F_X(e^y)$

So $F_Y(y)=F_X(e^y)$, then we differentiate to get $f_Y(y)=f_X(e^y)e^y$.

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Thankyou user65384 for your answer, but I am afraid to say that you did not get the point. I do not want to know how to get $F_Y(y)$, in my case I suppose I already have it. In your example you use $g(\cdot) = ln(\cdot)$ but in my question I pointed clear that $g$ must be $g(\cdot) =F_Y^{-1}(F_X(\cdot))$. In particular I need to consider the case where $F_Y(y)$ is a stair function that, so, cannot be inverted! – Andry Mar 20 '13 at 8:27

The transformation "almost" works if $X$ is continuous which implies that $F_X(x)$ has no jumps. Consider consecutive values $y'<y$ in support of $Y$. Extend $F^{-1}_Y(u)$ by letting it map the values from the interval $(P(Y\le y'),P(Y\le y)]=(F_Y(y'),F_Y(y)]$ to $y$. Note that the interval is half-open. If $y$ is the minimum value in the support of $Y$, then use $y'=-\infty$. The combined extended transformation then maps values of $X$ such that $F^{-1}_X(F_Y(y'))< X\le F^{-1}_X(F_Y(y))]$ to $y$.

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