Conditional Laws of Sums Given a Summand

Why Conditioning Simplifies a Sum

A sum $S=X+Y$ can have a complicated marginal distribution because both summands vary. Once $X$ is given, however, its contribution to the sum is fixed: the remaining randomness comes from $Y$, with its law conditioned on the observed value of $X$. This turns a problem about a sum into a translation of a conditional distribution.

The key point is that the conditional law of $Y$ may depend on $x$. Independence makes that law the same as the marginal law of $Y$; dependence generally does not. In either case, the distribution of $S$ given $X=x$ is obtained by adding $x$ to a draw from the conditional law of $Y$ given $X=x$.

The Translation Rule

Definition: Let $X$ and $Y$ be real-valued random variables, and let $K(x,B)$ be a regular conditional distribution of $Y$ given $X=x$. Thus, for each $x$, $K(x,\cdot)$ is a probability distribution on the real line, and $K(X,B)$ is a version of $P(Y\in B\mid X)$ for every Borel set $B$. The conditional distribution of $S=X+Y$ given $X=x$ is the probability distribution $$ K_S(x,A)=\int \mathbf{1}_{A}(x+y)\,K(x,dy) $$ for Borel sets $A$. Equivalently, it is the law of $x+Y$ when $Y$ has conditional law $K(x,\cdot)$.

For real-valued random variables, such conditional distributions exist. Their values as functions of $x$ are determined only up to sets of $X$-probability zero. If $X$ is discrete, conditioning at a value $x$ with $P(X=x)>0$ has its ordinary event-based meaning. If $X$ is continuous, the conditional law at a particular $x$ is understood through a regular conditional distribution; density formulas specify a useful version wherever the relevant marginal density is positive.

Theorem: Write $F_{Y\mid X=x}(u)=K(x,(-\infty,u])$. Then a version of the conditional CDF of $S=X+Y$ given $X=x$ is $$ F_{S\mid X=x}(t)=F_{Y\mid X=x}(t-x). $$ If $Y\mid X=x$ has a PMF $p_{Y\mid X}(y\mid x)$, then $$ p_{S\mid X}(s\mid x)=p_{Y\mid X}(s-x\mid x). $$ If $Y\mid X=x$ has a density $f_{Y\mid X}(y\mid x)$, then $$ f_{S\mid X}(s\mid x)=f_{Y\mid X}(s-x\mid x). $$ Each formula applies at values of $x$ where the stated conditional law is defined; in a density model, the conditional-density formula is ordinarily used where $f_X(x)>0$.

The CDF identity follows from the event equivalence $X+Y\le t$ if and only if $Y\le t-X$. For the PMF and density, the shift changes the argument but introduces no scale factor: the map from $y$ to $s=x+y$ has derivative one.

Why the Translation Rule Holds

Here is a measure-based justification that covers discrete, continuous, and mixed conditional laws. A regular conditional distribution $K$ has the integration property $$ E[h(X,Y)\mid X]=\int h(X,y)\,K(X,dy) $$ for every bounded Borel function $h$. This property follows first for indicator functions of rectangles from the definition of $K$, then for simple functions by linearity, and finally for bounded Borel functions by the standard monotone-class extension.

Take $h(x,y)=\mathbf{1}_A(x+y)$. The resulting conditional probability is $$ P(S\in A\mid X)=\int \mathbf{1}_A(X+y)\,K(X,dy)=K_S(X,A). $$ Thus $K_S$ is a version of the conditional distribution of $S$ given $X$. For the CDF, use $A=(-\infty,t]$. For each fixed $x$, $x+y\le t$ is equivalent to $y\le t-x$, which gives the claimed CDF formula.

For a conditional PMF, $S=s$ exactly when $Y=s-x$, so the mass at $s$ is the conditional mass at $s-x$. If the conditional law of $Y$ has density $f_{Y\mid X}(\cdot\mid x)$, then for Borel $A$, $$ P(x+Y\in A\mid X=x)=\int \mathbf{1}_A(x+y)f_{Y\mid X}(y\mid x)\,dy =\int_A f_{Y\mid X}(s-x\mid x)\,ds, $$ where the last equality uses $s=x+y$, so $ds=dy$. The integrand is therefore the conditional density of $S$. $\square$

Worked Applications

Worked Example: Independent Normal Summands

Suppose $X\sim N(2,1)$ and $Y\sim N(3,4)$ are independent, and let $S=X+Y$. Find the conditional distribution of $S$ given $X=x$, and calculate $P(S\le 8\mid X=4)$.

Independence means the conditional law of $Y$ given $X=x$ is its marginal law $N(3,4)$. Adding the fixed value $x$ shifts the mean by $x$, while leaving the variance unchanged. Therefore $$ S\mid X=x\sim N(x+3,4). $$

At $x=4$, the conditional mean is $7$ and the conditional variance is $4$, so the conditional standard deviation is $2$. Standardizing the threshold gives $$ P(S\le 8\mid X=4) =\Phi\left(\frac{8-7}{2}\right) =\Phi(0.5) \approx 0.6915. $$ The conditional variance is $4$, not $1+4$: once $X$ is given, $X$ contributes no remaining randomness.

Worked Example: Dependent Continuous Summands

Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)=2$ on $0<x<y<1$, and zero elsewhere. Find the conditional density and CDF of $S=X+Y$ given $X=x$, then compute $P(S\le 1.1\mid X=0.4)$.

First integrate the joint density over $y$ to find the marginal density of $X$: $$ f_X(x)=\int_x^1 2\,dy=2(1-x),\qquad 0<x<1. $$ Thus, for $0<x<1$ and $x<y<1$, $$ f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)} =\frac{2}{2(1-x)} =\frac{1}{1-x}. $$ So $Y\mid X=x$ is uniform on $(x,1)$. This distribution depends on $x$, reflecting the dependence between the summands.

Adding $x$ shifts the conditional support $(x,1)$ to $(2x,1+x)$. Hence $$ f_{S\mid X}(s\mid x)=\frac{1}{1-x},\qquad 2x<s<1+x, $$ and the conditional CDF is $$ F_{S\mid X=x}(t)= \begin{cases} 0, & t\le 2x,\\ \dfrac{t-2x}{1-x}, & 2x<t<1+x,\\ 1, & t\ge 1+x. \end{cases} $$ For $x=0.4$, the support is $(0.8,1.4)$. Since $1.1$ is inside it, $$ P(S\le 1.1\mid X=0.4) =\frac{1.1-0.8}{1.4-0.8} =\frac{0.3}{0.6} =0.5. $$

Worked Example: A Discrete Conditional Law

Let $P(X=0)=0.4$ and $P(X=1)=0.6$. Given $X=0$, suppose $Y$ is 0 with probability $0.75$ and 1 with probability $0.25$. Given $X=1$, suppose $Y$ is 1 with probability $0.2$ and 2 with probability $0.8$. Find the conditional PMF of $S=X+Y$ for each value of $X$.

When $X=0$, adding zero leaves the possible values unchanged: $$ P(S=0\mid X=0)=0.75,\qquad P(S=1\mid X=0)=0.25. $$ All other conditional masses are zero.

When $X=1$, add one to each possible value of $Y$: $$ P(S=2\mid X=1)=P(Y=1\mid X=1)=0.2, $$ $$ P(S=3\mid X=1)=P(Y=2\mid X=1)=0.8. $$ These masses sum to one in each case. The conditional support changes with $X$, so the conditional distribution cannot be described by simply shifting one common law for $Y$.

Conditional Moments of the Sum

The translation rule also gives useful moment identities. They hold without assuming independence: all dependence is already represented in the conditional moments of $Y$ given $X$.

Theorem: If $E[Y^2]<\infty$, then, almost surely, $$ E[S\mid X]=X+E[Y\mid X],\qquad \operatorname{Var}(S\mid X)=\operatorname{Var}(Y\mid X). $$

For the first identity, $S=X+Y$, and $X$ is measurable with respect to the information generated by $X$. Linearity and the defining property of conditional expectation give $$ E[S\mid X]=E[X+Y\mid X]=X+E[Y\mid X]. $$ For the variance identity, subtract this conditional mean from $S$: $$ S-E[S\mid X]=Y-E[Y\mid X]. $$ Taking the conditional expectation of the squares of both sides gives $$ \operatorname{Var}(S\mid X) =E[(S-E[S\mid X])^2\mid X] =E[(Y-E[Y\mid X])^2\mid X] =\operatorname{Var}(Y\mid X). $$ The finite second-moment assumption ensures these conditional variances are well-defined and finite almost surely. $\square$

Common Pitfall: Replacing a Conditional Law by a Marginal Law

A common mistake is to use the marginal distribution of $Y$ when finding the law of $S\mid X=x$, even though $X$ and $Y$ are dependent. For example, in the triangular-density example, the marginal law of $Y$ is not the uniform distribution on $(x,1)$ for every $x$. That interval describes the conditional support after $X=x$ is known, and its endpoint changes with $x$.

The correction is to identify $Y\mid X=x$ first, then shift that law by $x$. Independence is the special case in which $Y\mid X=x$ has the same distribution as $Y$; it is not a default assumption. For continuous $X$, also avoid interpreting $P(S\le t\mid X=x)$ as an ordinary ratio of probabilities: $P(X=x)$ is typically zero. Use a conditional distribution or an appropriate conditional-density formula instead.

Key takeaway: Given $X=x$, the sum $X+Y$ has the conditional law of $Y\mid X=x$ shifted by $x$. Shift both the values and the support, and retain any dependence through the conditional law.

Check Your Understanding

Use the conditional law of the remaining summand before shifting it.

If $Y\mid X=x$ is uniform on $(x,2x)$ for $x>0$, what is the support of $X+Y\mid X=x$?
State the conditional CDF of $X+Y$ given $X=x$ in terms of $F_{Y\mid X=x}$.
If $X$ and $Y$ are independent and $Y\sim N(1,9)$, what is the conditional distribution of $X+Y$ given $X=x$?
Why can the marginal density of $Y$ not generally be substituted for $f_{Y\mid X}(y\mid x)$?
Under a finite second-moment assumption, how does $\operatorname{Var}(X+Y\mid X)$ compare with $\operatorname{Var}(Y\mid X)$?