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I have two random variables $X$ and $Y$, specific elements that these random variables can take are $x$ and $y$. Now, say I define a random variable $f(x,Y)$ (a function of the random variable $Y$) as, $$f(x,Y)=g(x,Y)$$ Note here that $x$ is a specific value and $Y$ is the random variable. This is true for every $x$. I have a straightforward, and maybe silly doubt, is it true that $$f(X,y)=g(X,y)$$ Here I've flipped the random variable and the specific variable around. Intuitively, looking at random variables as functions, it should be true. Is there a more convincing argument?

Update: Since there was some confusion, it is a part of a bigger problem, where $$ f(x,Y)= \begin{cases} g(x,Y) &\text{if } E \\ h(x,Y) &\text{otherwise } \end{cases}$$ for some event $E$

arkarc
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  • The question is imprecise, as witnessed by the offtopic answers received so far, so let us try to reformulate it unambiguously and to answer it. $$ $$ Statement: If, for every $x$ in $X(\Omega)$ and every $\omega$ in $\Omega$, $f(x,Y(\omega))=g(x,Y(\omega))$ then indeed, for every $y$ in $Y(\Omega)$ and every $\omega$ in $\Omega$, $f(X(\omega),y)=g(X(\omega),y)$. $$ $$ Proof: Both conditions are equivalent to the fact that, for every $x$ in $X(\Omega)$ and every $y$ in $Y(\Omega)$, $f(x,y)=g(x,y)$. End of proof. – Did Oct 28 '15 at 09:07
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    Your first equality seems to indicate that $f$ and $g$ are the same function. –  Oct 28 '15 at 09:09
  • You definition of $f(x,Y)$ is peculiar. You are invoking a second function $g$ for that. That only works if $g$ is defined allready. But the equality $f(x,Y)=g(x,Y)$ then suggests that there is no need to define $f(x,Y)$. It equals $g(x,Y)$ which is allready defined. – drhab Oct 28 '15 at 09:10
  • @YvesDaoust Only on restricted sets of arguments (it seems), but yes. – Did Oct 28 '15 at 09:10
  • @did: it is not unlikely that the OP confuses random variable and "for any value", it you see what I mean. –  Oct 28 '15 at 09:12
  • Sorry if it's come off as imprecise. It's a part of a bigger problem, where $f(x,Y)$ can be two possible random variables $g(x,Y)$ or $h(x,Y)$ depending on an external event $E$. $$f(x,Y)= \begin{cases} g(x,Y) &\text{if } E \ h(x,Y) &\text{otherwise } \end{cases}$$ – arkarc Oct 28 '15 at 09:15
  • @YvesDaoust Indeed--which is why I explicited the arguments of the functions in my comment. We will see. (Ah, the last comment by the OP is bad news... :-)) – Did Oct 28 '15 at 09:15

2 Answers2

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Note that two functions can have the same value at the same points and be different. E.g.

\begin{align*} f(X, Y) = \frac{X}{\pi} + h(Y)\\ g(X, Y) = sin(X) + h(Y) \end{align*}

here, there are specific values of $x$ (specifically $\pi$) such that $f(x, Y) = g(x,Y)$, but $f(X,y) \neq g(X,y)$. If you have the condition that for all $x$ in $X$, and all $y$ in $Y$ then the equality should hold.

JDThinking
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If $X$ has values $0$ and $1$ and $Y$ takes values $3$ and $4$, then If $f(0,Y)=G(0,Y)$ means that $f(0,3)=g(0,3)$ and $f(0,4)=g(0,4)$. These relations do not imply that $f(1,3)=g(1,3)$ or $f(1,4)$ = $g(1,4)$.

kmitov
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