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I am having hard time interpreting the relationship, if any, between conditional probabilities vs. conditional probability distributions, in particular, regarding the number of random variables required to be able to define these two concepts.

As I understand, conditional probability can be defined for events involving a single random variable. For example, we could ask what is the probability that a die roll resulted in 4 given that the number is larger than 3, i.e., $P(X=4 | X>3)$.

However, the conditional probability distribution is based on the concept of joint distribution, which requires two or more random variables.

My question is whether it makes sense to talk about the conditional probability distribution of a single random variable or, conversely, whether a conditional probability statement like the one above can be related to a conditional probability distribution?

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  • $\begingroup$ Change $P(X=4|X>3)$. to $ P(X \le x|X>3)$ for (all) x, and now you have a conditional distribution function. Of course, this conditional distribution will be zero for all $x \le 3$. $\endgroup$ – Mark L. Stone Dec 8 '17 at 20:41
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    $\begingroup$ If your concept of conditional probabilities truly requires two random variables, then let $Y=X$ and define $\Pr(X=4\mid X\gt 3) = \Pr(X=4\mid Y \gt 3)$. $\endgroup$ – whuber Dec 8 '17 at 20:57
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    $\begingroup$ Or define a second random variable $Y=\mathbb{I}_{(3,\infty)}(X)$. $\endgroup$ – Xi'an Dec 8 '17 at 21:45
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    $\begingroup$ To get a better sense of where @Xi'an and I are coming from, see some of the better threads on conditional probability. They take various points of view, some of which do not need random variables at all. $\endgroup$ – whuber Dec 8 '17 at 22:19
  • $\begingroup$ Yes, I thought about defining a new variable $Y=X$ to end up with two random variables, but then $Y$ and $X$ become perfectly "correlated", which made it even harder for me to wrap my head around. $\endgroup$ – Kavka Dec 8 '17 at 22:41
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Well I think the term "conditional probability" or "conditional probability distribution" can both extend to two or more variables (correct me if I am wrong). For example, let us suppose we have $X,Y,Z$ are i.i.d random variables uniformly distributed on (0,1) for simplicity. Then we are required to find the conditional probability of $P(X \ge YZ|Y>\frac{1}{2})$. This should be a valid question about finding conditional probability.

Then,

$$P(X \ge YZ|Y>\frac{1}{2}) = \frac{P(X \ge YZ,Y>\frac{1}{2})}{P(Y>\frac{1}{2})} = \frac{\int_0^1\int_\frac{1}{2}^1\int_{yz}^1 1 dxdydz}{\frac{1}{2}}$$ $$= 2 \int_0^1\int_\frac{1}{2}^1(1-yz)dydz = 2\int_0^1 \Big(\frac{1}{2}-\frac{3z}{8}\Big)dz=2 \Big( \frac{1}{2}-\frac{3}{16}\Big)=\frac{5}{8}$$

Similarly, $$P(X \ge YZ|Y \le\frac{1}{2})= \frac{\int_0^1\int_0^\frac{1}{2}\int_{yz}^1 1 dxdydz}{\frac{1}{2}} = \frac{7}{8}$$ , as you can verify.

In a nutshell, the concept of conditional probability is not only valid on single variable case.

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    $\begingroup$ I think you meant to say $Y < 1/2$ in your latter example. $\endgroup$ – Kavka Dec 14 '17 at 16:49
  • $\begingroup$ My question is the converse of the question that you've answered: "...whether it makes sense to talk about the conditional probability distribution of a single random variable", the same way we can do for conditional probabilities. $\endgroup$ – Kavka Dec 14 '17 at 16:52
  • $\begingroup$ Thanks Kavka~ Yes I meant for $Y< \frac{1}{2}$. Oh, I got it. Then you may consider the exponential or geometric distribution and the memoryless property. For exponential distribution, for example, you have $P(X \ge s+t | X \ge s) = P(X \ge t)$, for $t >0, s>0$. $\endgroup$ – son520804 Dec 14 '17 at 17:09

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