Conditional Support and Normalization

Why Conditional Support Changes

A joint distribution restricts pairs of values \((x,y)\) to a joint support. Once \(Y=y\) is observed, the possible values of \(X\) are no longer all the values allowed by its marginal distribution: they are the values compatible with that particular slice of the joint support. If the slice changes as \(y\) changes, the conditional support changes too.

This matters directly for normalization. A conditional pmf must sum to one over the values of \(X\) allowed at the observed \(y\); a conditional density must integrate to one over the corresponding interval or region. Using the unconditional support, or bounds that do not match the slice, can produce a formula that is not a probability distribution.

Definition: For discrete random variables \(X\) and \(Y\), the conditional support of \(X\) given \(Y=y\), when \(P(Y=y)>0\), is the set of \(x\) values for which \(P(X=x,Y=y)>0\). For a joint density \(f_{X,Y}(x,y)\), the conditional support at a value \(y\) with marginal density \(f_Y(y)>0\) is given, up to sets of \(x\)-values of Lebesgue measure zero, by the \(x\)-values for which \(f_{X,Y}(x,y)>0\).

For a density, “support” needs this qualification because changing a density on a set of measure zero does not change the distribution. In particular, including or excluding an endpoint usually makes no difference to a continuous probability. In examples, inequalities describe the region where the density is positive; interval endpoints can be included or excluded without changing the integrals.

Conditional Formulas on a Slice

For a discrete joint distribution, write \(p_{X,Y}(x,y)=P(X=x,Y=y)\) and \(p_Y(y)=P(Y=y)\). For any \(y\) with \(p_Y(y)>0\), the conditional pmf is the joint mass divided by the marginal mass. It is zero outside the slice where the joint mass is positive.

p_{X\mid Y}(x\mid y) = \frac{p_{X,Y}(x,y)}{p_Y(y)}, \qquad p_Y(y)=\sum_x p_{X,Y}(x,y).

For a joint density, define the marginal density by integrating along the slice at \(y\). Where \(f_Y(y)>0\), the conditional density is the joint density divided by this marginal density.

f_Y(y)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx, \qquad f_{X\mid Y}(x\mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}.

These formulas show how to find the conditional support: hold \(y\) fixed, identify the allowed \(x\)-values in the joint support, and use only those values in the conditional formula. If the joint density has a factor depending only on \(y\), that factor cancels when the numerator is divided by \(f_Y(y)\). The resulting conditional density still needs to be checked on the \(x\)-range at that \(y\).

Normalization Follows from the Marginal

The central check is that the conditional probabilities add or integrate to one on the slice. The statements below make the role of the support explicit.

Theorem: If \(X,Y\) are discrete and \(p_Y(y)>0\), then \(\sum_{x:p_{X,Y}(x,y)>0}p_{X\mid Y}(x\mid y)=1\). If \((X,Y)\) has a joint density and \(f_Y(y)>0\), then \(\int_{\{x:f_{X,Y}(x,y)>0\}}f_{X\mid Y}(x\mid y)\,dx=1\), with density statements understood up to sets of measure zero.

Proof for the discrete case. Since \(p_Y(y)>0\), at least one joint mass in the slice is positive. Terms outside the slice have joint mass zero, so summing over the slice gives

\sum_{x:p_{X,Y}(x,y)>0}p_{X\mid Y}(x\mid y) = \frac{\sum_{x:p_{X,Y}(x,y)>0}p_{X,Y}(x,y)}{p_Y(y)} = \frac{p_Y(y)}{p_Y(y)} =1.

Proof for the density case. On the slice, divide the joint density by the positive constant \(f_Y(y)\). The joint density is zero outside that slice, up to null sets. Therefore

\int_{\{x:f_{X,Y}(x,y)>0\}} f_{X\mid Y}(x\mid y)\,dx = \frac{1}{f_Y(y)} \int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx = \frac{f_Y(y)}{f_Y(y)} =1.

The integral equality uses the definition of \(f_Y(y)\); the integrand contributes zero off the slice. A value \(y\) with zero marginal mass in the discrete case, or zero marginal density in the continuous case, cannot be handled by these ratios. The formulas here apply only when the denominator is positive.

Worked Example: A Discrete Slice with a Growing Support

Suppose \((X,Y)\) is uniform on the six pairs \((x,y)\) satisfying \(1\leq x\leq y\leq 3\), where \(x\) and \(y\) are integers. Thus each allowed pair has joint mass \(1/6\). Find the support and conditional pmf of \(X\) given each possible \(Y=y\).

For a fixed \(y\), the joint support allows \(x=1,\ldots,y\). There are \(y\) such pairs, so \(p_Y(y)=y(1/6)=y/6\). For each allowed \(x\),

p_{X\mid Y}(x\mid y) = \frac{1/6}{y/6} = \frac{1}{y}, \qquad x\in\{1,\ldots,y\}.

When \(y=1\), the conditional support is \(\{1\}\), and its only probability is \(1\). When \(y=2\), the support is \(\{1,2\}\), with probability \(1/2\) at each value. When \(y=3\), the support is \(\{1,2,3\}\), with probability \(1/3\) at each value. In each case the probabilities sum to one. The support expands with \(y\), rather than remaining equal to the full marginal support \(\{1,2,3\}\).

Moving Bounds for Conditional Densities

For a continuous joint distribution, fixing \(y\) can turn a two-dimensional support region into an interval whose endpoints depend on \(y\). To normalize the conditional density, integrate over that interval, not over a fixed range copied from the marginal support.

Worked Example: A Triangular Support Region

Let the joint density be \(f_{X,Y}(x,y)=2\) when \(0<y<x<1\), and zero elsewhere. Find the conditional density of \(X\) given \(Y=y\), and calculate \(P(0.6<X<0.8\mid Y=0.4)\).

For a fixed \(y\) with \(0<y<1\), the allowed \(x\)-values satisfy \(y<x<1\). Integrating the joint density over this slice gives

f_Y(y)=\int_y^1 2\,dx=2(1-y).

Thus the conditional density is

f_{X\mid Y}(x\mid y) = \frac{2}{2(1-y)} = \frac{1}{1-y}, \qquad y<x<1.

Its normalization checks directly: \(\int_y^1 \frac{1}{1-y}\,dx=\frac{1-y}{1-y}=1\). At \(y=0.4\), the conditional support is \(0.4<x<1\) and the density is \(1/0.6=5/3\). The requested interval lies within that support, so

P(0.6<X<0.8\mid Y=0.4) = \int_{0.6}^{0.8}\frac{5}{3}\,dx = \frac{5}{3}(0.2) = \frac{1}{3}.

The lower bound of the conditional support is \(y\), not zero. Using zero as the lower bound would include \(x\)-values that cannot occur jointly with the observed \(Y=y\).

Worked Example: Both Conditional Bounds Depend on the Observation

Let \(f_{X,Y}(x,y)=2\) for \(0<y<1\) and \(y<x<2y\), and let it be zero elsewhere. Find \(f_{X\mid Y}(x\mid y)\), verify its normalization, and compute \(P(0.36<X<0.51\mid Y=0.3)\).

At a fixed \(y\in(0,1)\), the conditional support is the interval \(y<x<2y\). Its length is \(2y-y=y\), so the marginal density of \(Y\) is

f_Y(y)=\int_y^{2y}2\,dx=2y.

Dividing by this marginal density gives

f_{X\mid Y}(x\mid y) = \frac{2}{2y} = \frac{1}{y}, \qquad y<x<2y.

The integral over the conditional support is \(\int_y^{2y}(1/y)\,dx=(2y-y)/y=1\). When \(y=0.3\), the support is \(0.3<x<0.6\), and the conditional density is \(1/0.3=10/3\). The interval \((0.36,0.51)\) is inside the support, giving

P(0.36<X<0.51\mid Y=0.3) = \int_{0.36}^{0.51}\frac{10}{3}\,dx = \frac{10}{3}(0.15) = 0.5.

Here both endpoints of the conditional support move with \(y\). The interval length also changes with \(y\), which is why the normalizing constant \(1/y\) changes.

Common Mistake: Normalizing Over the Wrong Range

A frequent error is to identify the marginal support of \(X\) and then use it unchanged after conditioning. In the first continuous example, the marginal support of \(X\) includes values in \((0,1)\), but after observing \(Y=y\), only \(y<x<1\) is possible. The conditional density is \(1/(1-y)\) on that slice, not a constant normalized over all of \((0,1)\).

Another error is to write a plausible conditional formula and skip its normalization check. For instance, in the second continuous example, the joint density has the constant value 2. That does not make the conditional density equal to 2: the conditional formula must divide by \(f_Y(y)=2y\), producing \(1/y\) on the moving interval. Integrating this conditional density over the slice confirms that its total probability is one.

Key takeaway: Fix the observed value first, read the allowable values from the corresponding slice of the joint support, and normalize on that slice. For a pmf, the slice masses sum to the marginal mass; for a density, the slice integral is the marginal density.

Check Your Understanding

For each question, identify the conditional support before attempting any normalization.

A discrete joint pmf is positive only at integer pairs with \(0\leq x\leq y\leq 4\). What is the support of \(X\) given \(Y=2\)?
If \(p_{X,Y}(x,y)\) is positive at three values of \(x\) for a fixed \(y\), what sum must the conditional probabilities over those values equal?
A joint density is positive for \(0<y<x<2\). For a fixed \(y\in(0,2)\), state the conditional support of \(X\), and write the integral used to find \(f_Y(y)\).
In the density \(f_{X,Y}(x,y)=2\) on \(0<y<1,\ y<x<2y\), what is the conditional support when \(y=0.6\)?
Why can the ratio formulas for a conditional pmf or density not be used when the marginal denominator is zero?