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Assume we are given a joint distribution $P(X,Y)$ where $P(0,0)=0.1$, $P(0,1) = 0.4$, $P(1,0)=0.3$, and $P(1,1)=0.2$. The goal is to compute $P(X|Y=1)$.

Traditionally, solving a conditional probability problem $P(A|B)$ simplifies to $\frac{P(A,B)}{P(B)}$, but I'm unsure how to apply it to this case. In particular, I am unclear on what the probability $P(X,Y=1)$ means since $P(X)$ is a marginal probability.

To avoid this, I enumerated the different values $X$ takes on and plugged it into the original quantity to solve -- $P(X|Y=1)$:

  • $P(X=0|Y=1) = 0.4/0.6 = 4/6$
  • $P(X=1|Y=1) = 0.2/0.6 = 2/6$

This gives me the final answer of $P(X|Y=1) = [4/6, 2/6]$, but I'm not sure whether the answer should be multiple probabilities or a single probability.

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  • $\begingroup$ P(X|Y=2) is a distribution in the same way that P(X) is also a distribution (and not just one "probability"). One probably is P(X=0, Y=0)=0.1, for example. $\endgroup$ – nbro Jan 28 at 2:14
  • $\begingroup$ @nbro Does this mean my answer is correct? $\endgroup$ – Shrey Jan 28 at 4:04
  • $\begingroup$ you're ok. just as $p(x,y)$ is a 2D function(table), $p(x)$, $p(x|Y=y)$ are 1D functions (rows). $\endgroup$ – gunes Jan 28 at 4:52
  • $\begingroup$ $P(X|Y=1)$ is shorthand for $P(X=x|Y=1)$ $\endgroup$ – StatsStudent Jan 28 at 5:47
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You have gone about this correctly, but the final answers are typically written as a function of $x$. It's helpful, I think, to remember that $P(X|Y=1)$ is just shorthand for $P(X=x|Y=1)$ where $x$ is any number in the support of $x$ (in this case 0 and 1). So you'd calculate this as follows:

\begin{eqnarray*} \\{P(X|Y=1)} & = & {P\left(X=x|Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{P\left(Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{P\left(X=0,\,Y=1\right)+P\left(X=1,\,Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{0.4+0.2}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{0.6} \end{eqnarray*}

Now, writing this as a function of $x$ gives:

\begin{eqnarray*} P\left(X=x|Y=1\right) & = & \begin{cases} \frac{P\left(X=0,\,Y=1\right)}{0.6} & ,\text{for }x=0\\ \frac{P\left(X=1,\,Y=1\right)}{0.6} & ,\text{for }x=1 \end{cases}\\ & = & \begin{cases} \frac{0.4}{0.6} & ,\text{for }x=0\\ \frac{0.2}{0.6} & ,\text{for }x=1 \end{cases}\\ & = & \begin{cases} \frac{2}{3} & ,\text{for }x=0\\ \frac{1}{3} & ,\text{for }x=1 \end{cases} \end{eqnarray*}

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You are correct. You are supposed to be getting "multiple" probabilities.

You said that your goal is to figure out $P(X|Y = 1)$. Well, in order to achieve your goal, you need to know what the probability that $X = 0$ given that $Y = 1$ is as well as what the probability that $X = 1$ given that $Y = 1$ is, which you did.

nbro's comment is basically telling you that when you get "multiple" probabilities, you are actually computing a probability mass (or density) function (or, equivalently, yet not the same, the distribution of the random variable $X$ given $Y = 1$).

Note: $P(X|Y=1)=[4/6,2/6]$ is the probability mass function of the random variable $X$ given $Y = 1$ because it tells you what that probability is for all the possible values of $X$, ie, $X = 0$ and $X$ = 1.

Finally, I use "multiple" in quotes because nobody really says that. In your case, they would ask something like: "This gives me the final answer of P(X|Y=1)=[4/6,2/6], but I'm not sure whether the answer should be ${\it \text{ a probability mass function}}$ or a single probability."

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