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I am taking a beginner's probability theory course. One thing that is confusing me a lot. We know that the probability of a continuous random variable at a fixed point is $0$. But when learning about conditional density functions in joint distributions, we know that

For any $y_2$ such that $f_2(y_2) > 0$, the conditional density of $Y_1$ given $Y_2 = y_2$ is given by $$f (y_1\mid y_2) = \frac{f (y_1, y_2)}{f_2(y_2)}$$ If we say for instance the probability of $Y_2$ at a fixed point is $0$, how does it make sense to take a fixed value of $Y_2$ when using conditional density function?

Please don't mark this as duplicate, I have seen all other answers on conditional probability, but I am looking for an intuitive answer possibly with an intuitive example to clear up the confusion. All other answers state we can condition on events with zero probability in continuous distributions, but none of them state why.

Can someone please explain, thanks.

Henry
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Ali Ahmed
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  • Here's an intuitive example. Suppose $5$ people enter an elevator with capacity $1200 \text{lbs}$; take $S$ as the combined weight of all five people. Let $X=1$ if the elevator's weight capacity is exceeded and $X=0$ otherwise. Note $\mathbb{P}(S=1000)=0$ but we can still interpret an expression like $f_{X|S}(1|1000)$. Do you know what this equals? –  Feb 26 '23 at 21:04
  • @MatthewH. I think this question may be about densities rather than probabilities – Henry Feb 26 '23 at 21:42
  • @Henry you could modify my example to have it relate to conditional densities as follows. For example, suppose we randomly select an individual from the world. Let $X$ be the age (in years) and $Y$ the weight (in pounds) of the individual. Both $X,Y$ are continuous random variables, so $\mathbb{P}(X=0.235)=0$, and a probability like $\mathbb{P}(Y\geq 13|X=0.235)$ is still a meaningful expression which may be evaluated using conditional densities. –  Feb 26 '23 at 21:52
  • Does this mean since we are taking ratio of densities and not probabilities in conditional density, we can have a fixed value? I get the expression but the same problem is like you said if $P(X=0.234) = 0$ then it would mean in "given that" it would also have a chance of occuring $0$. – Ali Ahmed Feb 27 '23 at 05:03
  • A more rigorous definition of the density of a random variable $X$ conditioned on ${Y=y}$, with $X$ and $Y$ being absolutely continuous is $$ f_{X\mid Y}(x\mid y) = \lim_{\Delta x\to 0,\Delta y\to 0}=\frac{\frac1{\Delta x}\mathbb P(x<X\leqslant x+\Delta x)}{\mathbb P(y<Y\leqslant y+\Delta y)}. $$ – Math1000 Feb 28 '23 at 13:27

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