• The conditional distribution of $$Y$$ given $$X=x$$ is the distribution of $$Y$$ values over only those outcomes for which $$X=x$$. It is a distribution on values of $$Y$$ only; treat $$x$$ as a fixed constant when conditioning on the event $$\{X=x\}$$.
• Conditional distributions can be obtained from a joint distribution by slicing and renormalizing.
• Conditioning on the value of a random variable involves treating that random variable as a constant.
• It is sometimes possible to identify one-way conditional distributions ($$Y$$ given $$X$$, or $$X$$ given $$Y$$) simply by inspecting the joint distribution, without doing any calculations.
• Be sure to distinguish between joint, conditional, and marginal distributions.
• The joint distribution is a distribution on $$(X, Y)$$ pairs. A mathematical expression of a joint distribution is a function of both values of $$X$$ and values of $$Y$$. Pay special attention to the possible values; the possible values of one variable might be restricted by the value of the other.
• The conditional distribution of $$Y$$ given $$X=x$$ is a distribution on $$Y$$ values (among $$(X, Y)$$ pairs with a fixed value of $$X=x$$). A mathematical expression of a conditional distribution will involve both $$x$$ and $$y$$, but $$x$$ is treated like a fixed constant and $$y$$ is treated as the variable. Note: the possible values of $$Y$$ might depend on the value of $$x$$, but $$x$$ is treated like a constant.
• The marginal distribution of $$Y$$ is a distribution on $$Y$$ values only, regardless of the value of $$X$$. A mathematical expression of a marginal distribution will have only values of the single variable in it; for example, an expression for the marginal distribution of $$Y$$ will only have $$y$$ in it (no $$x$$, not even in the possible values).

## 23.1 Discrete random variables: Conditional probability mass functions

Example 23.1 Roll a fair four-sided die once and let $$X$$ be the number rolled. Then flip a fair coin $$X$$ times and let $$Y$$ be the number of heads.

1. Identify the possible values of $$X$$.

2. Identify the possible values of $$Y$$.

3. Find the conditional distribution of $$Y$$ given $$X=4$$.

4. Find the conditional distribution of $$Y$$ given $$X=3$$.

5. Find the probability that $$X=3$$ and $$Y=2$$.

6. Find the probability that $$X=3$$ and $$Y=y$$ for $$y = 0, 1, 2, 3, 4$$.

7. Find the joint distribution of $$X$$ and $$Y$$.

8. Find the marginal distribution of $$Y$$.

9. Find the conditional distribution of $$X$$ given $$Y=2$$.

• Let $$X$$ and $$Y$$ be two discrete random variables defined on a probability space with probability measure $$\text{P}$$. For any fixed $$x$$ with $$\text{P}(X=x)>0$$, the conditional probability mass function (pmf) of $$Y$$ given $$X=x$$ is a function $$p_{Y|X}$$ defined by $$p_{Y|X}(y|x)=\text{P}(Y=y|X=x)$$. \begin{align*} p_{Y|X}(y|x) = \text{P}(Y=y|X=x) & = \frac{\text{P}(X=x,Y=y)}{\text{P}(X=x)} = \frac{p_{X,Y}(x,y)}{p_X(x)}& & \text{a function of y for fixed x} \end{align*}
• To emphasize, the notation $$p_{Y|X}(\cdot|x)$$ represents the distribution of the random variable $$Y$$ given a fixed value $$x$$ of the random variable $$X$$. In the expression $$p_{Y|X}(y|x)$$, $$y$$ is treated as the variable and $$x$$ is treated like a fixed constant.
• Notice that the pmfs satisfy $\text{conditional} = \frac{\text{joint}}{\text{marginal}}$
• Conditional distributions can be obtained from a joint distribution by slicing and renormalizing. The conditional pmf of $$Y$$ given $$X=x$$ can be thought of as:
• the slice of the joint pmf $$p_{X, Y}(x, y)$$ of $$(X, Y)$$ corresponding to $$X=x$$, a function of $$y$$ alone,
• renormalized — by dividing by $$p_X(x)$$ — so that the probabilitiess, corresponding to different $$y$$ values, for the slice sum to 1.
• For a fixed $$x$$, the shape of the conditional pmf of $$Y$$ given $$X=x$$ is determined by the shape of the $$x$$-slice of the joint pmf, $$p_{X, Y}(x, y)$$. That is, $\text{As a function of values of Y}, \quad p_{Y|X}(y|x) \propto p_{X, Y}(x, y)$
• For each fixed $$x$$, the conditional pmf $$p_{Y|X}(\cdot |x)$$ is a different distribution on values of the random variable $$Y$$. There is not one “conditional distribution of $$Y$$ given $$X$$”, but rather a family of conditional distributions of $$Y$$ given different values of $$X$$.
• Rearranging the definition of a conditional pmf yields the multiplication rule for pmfs of discrete random variables \begin{align*} p_{X,Y}(x,y) & = p_{Y|X}(y|x)p_X(x)\\ & = p_{X|Y}(x|y)p_Y(y)\\ \text{joint} & = \text{conditional}\times\text{marginal} \end{align*}
• Marginal distributions can be obtained from the joint distribution by collapsing/stacking using the law of total probability. The law of total probability for pmfs is \begin{align*} p_{Y}(y) & = \sum_x p_{X,Y}(x, y)\\ & =\sum_x p_{Y|X}(y|x)p_X(x) \end{align*}

## 23.2 Continuous random variables: Conditional probability density functions

• Let $$X$$ and $$Y$$ be two continuous random variables with joint pdf $$f_{X,Y}$$ and marginal pdfs $$f_X, f_Y$$. For any fixed $$x$$ with $$f_X(x)>0$$, the conditional probability density function (pdf) of $$Y$$ given $$X=x$$ is a function $$f_{Y|X}$$ defined by \begin{align*} f_{Y|X}(y|x) &= \frac{f_{X,Y}(x,y)}{f_X(x)}& & \text{a function of y for fixed x} \end{align*}
• To emphasize, the notation $$f_{Y|X}(y|x)$$ represents a conditional distribution of the random variable $$Y$$ for a fixed value $$x$$ of the random variable $$X$$. In the expression $$f_{Y|X}(y|x)$$, $$x$$ is treated like a constant and $$y$$ is treated as the variable.
• Notice that the pdfs satisfy $\text{conditional} = \frac{\text{joint}}{\text{marginal}}$
• Conditional distributions can be obtained from a joint distribution by slicing and renormalizing. The conditional pdf of $$Y$$ given $$X=x$$ can be thought of as:
• the slice of the joint pdf $$f_{X, Y}(x, y)$$ of $$(X, Y)$$ corresponding to $$X=x$$, a function of $$y$$ alone,
• renormalized — by dividing by $$f_X(x)$$ — so that the density heights, corresponding to different $$y$$ values, for the slice are such that the total area under the density slice is 1.
• For a fixed $$x$$, the shape of the conditional pdf of $$Y$$ given $$X=x$$ is determined by the shape of the $$x$$-slice of the joint pdf, $$f_{X, Y}(x, y)$$. That is, $\text{As a function of values of Y}, \quad f_{Y|X}(y|x) \propto f_{X, Y}(x, y)$
• For each fixed $$x$$, the conditional pdf $$f_{Y|X}(\cdot |x)$$ is a different distribution on values of the random variable $$Y$$. There is not one “conditional distribution of $$Y$$ given $$X$$”, but rather a family of conditional distributions of $$Y$$ given different values of $$X$$.
• Rearranging the definition of a conditional pdf yields the multiplication rule for pdfs of continuous random variables \begin{align*} f_{X,Y}(x,y) & = f_{Y|X}(y|x)f_X(x)\\ & = f_{X|Y}(x|y)f_Y(y)\\ \text{joint} & = \text{conditional}\times\text{marginal} \end{align*}
• Marginal distributions can be obtained from the joint distribution by collapsing/stacking using the law of total probability. The law of total probability for pmfs is \begin{align*} f_{Y}(y) & = \int_{-\infty}^\infty f_{X,Y}(x, y)\, dx\\ & =\int_{-\infty}^\infty f_{Y|X}(y|x)f_X(x)\, dx \end{align*}
• Remember that the probability that a continuous random variable is equal to a particular value is 0; that is, for continuous $$X$$, $$\text{P}(X=x)=0$$. When we condition on $$\{X=x\}$$ we are really conditioning on $$\{|X-x|<\epsilon\}$$ and seeing what happens in the idealized limit when $$\epsilon\to0$$.
• When simulating, never condition on $$\{X=x\}$$; rather, condition on $$\{|X-x|<\epsilon\}$$ where $$\epsilon$$ represents some suitable degree of precision (e.g. $$\epsilon=0.005$$ if rounding to two decimal places).
• Remember pdfs do not return probabilities directly; $$f_{Y|X}(y|x)$$ is not a probability of anything. But $$f_{Y|X}(y|x)$$ is related to the probability that $$Y$$ is “close to” $$y$$ given that $$X$$ is “close to” $$x$$: $\text{P}(y-\epsilon/2<Y < y+\epsilon/2\; \vert\; x-\epsilon/2<X < x+\epsilon/2) \approx \epsilon f_{Y|X}(y|x)$

Example 23.2 Recall the continuous analog of the four-sided die problem. Spin the Uniform(1, 4) spinner twice and let $$X$$ be the sum of the two spins and $$Y$$ the larger to the two spins (or the common value if a tie). Recall that the joint pdf is

$f_{X, Y}(x, y) = \begin{cases} 2/9, & 2<x<8,\; 1<y<4,\; x/2<y<x-1,\\ 0, & \text{otherwise,} \end{cases}$ the marginal pdf of $$Y$$ is $f_Y(y) = \begin{cases} (2/9)(y-1), & 1<y<4,\\ 0, & \text{otherwise,} \end{cases}$ and the marginal pdf of $$X$$ is $f_X(x) = \begin{cases} (1/9)(x-2), & 2 < x< 5,\\ (1/9)(8-x), & 5<x<8,\\ 0, & \text{otherwise.} \end{cases}$

1. Find $$f_{X|Y}(\cdot|3)$$, the conditional pdf of $$X$$ given $$Y=3$$.

2. Find $$\text{P}(X > 5.5 | Y = 3)$$.

3. Find $$f_{X|Y}(\cdot|4)$$, the conditional pdf of $$X$$ given $$Y=4$$.

4. Find $$\text{P}(X > 5.5 | Y = 4)$$.

5. Find $$f_{X|Y}(\cdot|y)$$, the conditional pdf of $$X$$ given $$Y=y$$, for $$1<y<4$$.

6. Find $$f_{Y|X}(\cdot|3.5)$$, the conditional pdf of $$Y$$ given $$x=3.5$$.

7. Find $$f_{Y|X}(\cdot|6)$$, the conditional pdf of $$Y$$ given $$x=6$$.

8. Find $$f_{Y|X}(\cdot|x)$$, the conditional pdf of $$Y$$ given $$x$$.

Example 23.3 Suppose $$X$$ and $$Y$$ are continuous RVs with joint pdf

$f_{X, Y}(x, y) = \frac{1}{x}e^{-x}, \qquad x > 0,\quad 0<y<x.$

1. Identify by name the one-way conditional distributions that you can obtain from the joint pdf, without doing any calculus or computation.

2. Identify by name the marginal distribution you can obtain without doing any calculus or computation.

3. Describe how could you use the Exponential(1) spinner and the Uniform(0, 1) spinner to generate an $$(X, Y)$$ pair.

4. Sketch a plot of the joint pdf.

5. Sketch a plot of the marginal pdf of $$Y$$.

6. Set up the calculation you would perform to find the marginal pdf of $$Y$$.