Joint PMF of two random variables when one of them is replaced by its invertible function

Question

Say we have two discrete random variables $X_1$ and $X_2$ with respective ranges $R_{X_1}$ and $R_{X_2}$. Let $Y_1 = g(X_1)$, where $g$ is an invertible function (that is, for every $y \in R_{Y_1}$, there is exactly one $x \in R_{X_1}$ such that $y = g(x)$).

Now my question is whether the following is true: $$P\left(X_1=x_1,X_2=x_2\right) = P\left(Y_1=g(x_1),X_2=x_2\right)$$

P.S.: I think I can show that this is true if $X_1$ and $X_2$ are independent as follows: \begin{align}P\left(X_1=x_1,X_2=x_2\right) &= P(X_1=x_1)\cdot P(X_2=x_2)\\&=P(Y_1=g(x_1))\cdot P(X_2=x_2)\\&=P(Y_1=g(x_1),X_2=x_2)\end{align} where the last line follows from the fact that $Y_1$ and $X_2$ are independent (since functions of independent random variables are also independent).

But what about the case when $X_1$ and $X_2$ are not independent?

Answer 1

What you wanted to show is that when $X_1$ and $X_2$ are independent, so are $Y_1$ and $X_2$.

For all supported $y,z$:

$\qquad\begin{align}\mathsf P(Y_1\,{ =}\, y, X_2\,{ =}\,z) &= \mathsf P\bigl(X_1\,{ =}\,g^{-1}(y), X_2\,{ =}\,z\bigr)&&\ni g\text{ is bijective (over the R.V. supports)}\\&=\mathsf P\bigl(X_1\,{ =}\,g^{-1}(y)\bigr)\cdot\mathsf P(X_2\,{ =}\,z)&&\text{ by independence}\\&=\mathsf P(Y_1\,{ =}\,y)\cdot\mathsf P(X_2\,{ =}\,z)\end{align}$

Which was not quite what you had.

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Now, of course, when $X_1, X_2$ are not independent, then neither shall $Y_1, X_2$ be so.

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However

Now my question is whether the following is true: $$P\left(X_1=x_1,X_2=x_2\right) = P\left(Y_1=g(x_1),X_2=x_2\right)$$

Yes. That is immediate, since by definition of $Y_1=g(X_1)$.

Just that. We do not need independence for this.