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I was just solving problems on conditional probabilities. When I was taught this concept, most of the conditions were given for specific value. But then I saw an example where the condition was that X = 3Z where Z is a random variable. I do not understand this. For example, I can evaluate (1) easily but (2) seems tough.

(1) \begin{align*} \Pr(X+Y = k \mid Y=y) &= \frac{\Pr(X+Y = k \cap Y=y)}{\Pr(Y=y)} \\ &= \frac{\Pr(X = k-y \cap Y=y)}{\Pr(Y=y)} \\ &= \Pr(X = k-y \mid Y=y). \end{align*}

(2) \begin{align*} \Pr(X+Y = k \mid Y=3Z) \end{align*}

I tried searching online but it was mainly answered using $\sigma$-algebra. Could someone please show how would I evaluate (2), as in can I substitute the Y with 3Z? or what it means intuitively?

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    $\begingroup$ Do you have any clue about the joint distribution of $(X,Y,Z)$? In particular, are some of them independent? $\endgroup$ Commented Nov 27 at 19:52
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    $\begingroup$ The issue is that $Y = 3Z$ is not an event that one can condition on in a natural way. $Y=y$ is (arguably) more natural to condition on. If $P(Y=3Z)>0$, then just use the definition of conditional probability that you know. $\endgroup$ Commented Nov 27 at 20:31
  • $\begingroup$ It is worth considering what you mean by 'evaluate'. In the case of i), you have used two processes to simplify a term. What is your aim with ii)? $\endgroup$ Commented Nov 27 at 21:03

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$$\Pr(X+Y = k \mid Y=3Z) = \dfrac{\Pr(X+Y = k \cap Y=3Z)}{\Pr(Y=3Z)}= \dfrac{\Pr(X+3Z = k \cap Y=3Z)}{\Pr(Y=3Z)}$$ may give you the answer if you know the joint distribution of $(X,Y,Z)$ and if $\Pr(Y=3Z)>0$.

If $\Pr(Y=3Z)=0$ then you cannot divide by it and you could face the Borel–Kolmogorov paradox. For example if you use limits to try to avoid division by $0$ then you may find $$\lim\limits_{\varepsilon \to 0^+} \Pr(X+Y = k \mid -\varepsilon < Y-3Z < \varepsilon) \not = \lim\limits_{\varepsilon \to 0^+} \Pr(X+Y = k \mid 3-\varepsilon < \tfrac{Y}{Z} < 3+\varepsilon).$$

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