Worked Problems on Distribution of Sums

Introduction

A sum distribution is not found by applying one universal rule. The appropriate method depends on the distributions of the summands, their supports, and whether they are independent. A discrete sum may require combining several pairs of outcomes; a continuous sum may require an integral whose bounds change with the value of the sum; and some familiar distribution families have direct sum rules.

This tutorial works through those choices in detail. In each problem, first state the model and its assumptions, then choose a method, carry out the calculation, and check that the result is consistent. The techniques are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: If $X$ and $Y$ are random variables, their sum is $S=X+Y$. When $X$ and $Y$ are independent and discrete, the probability of a possible sum $s$ is $$ P(S=s)=\sum_x P(X=x)P(Y=s-x), $$ where the sum includes values of $x$ for which both probabilities are defined. When they are independent and have densities, the density of the sum is $$ f_S(s)=\int_{-\infty}^{\infty} f_X(x)f_Y(s-x)\,dx. $$ In either case, the support of the summands determines which terms or integration bounds contribute.

Choose the Method From the Model

The methods below are useful when their assumptions fit the problem. Independence is central to the convolution formulas and to the additivity results for Poisson and normal variables. If summands are dependent, those formulas cannot automatically be used; the joint distribution or a valid conditioning argument is needed.

Model and assumptions	Useful method	Check before calculating
Independent discrete variables	Combine pairs of outcomes that yield the same sum	Include every contributing pair exactly once
Independent continuous variables with densities	Convolution integral	Find the overlap of the supports for each value of the sum
Independent normal variables and a weighted sum	Use the linear-combination normal result	Multiply each variance by the square of its coefficient
Independent Poisson variables	Add the Poisson parameters	Check independence and then calculate the requested probability

A short derivation explains the discrete formula. If $S=X+Y=s$, then for each possible value $x$ of $X$, the only compatible value of $Y$ is $s-x$. These disjoint possibilities partition the event $\{S=s\}$. Therefore,

\begin{aligned} P(S=s) &=\sum_x P(X=x,\;Y=s-x)\\ &=\sum_x P(X=x)P(Y=s-x), \end{aligned}

where the second equality uses independence. For continuous independent variables, the same idea is expressed through the joint density: set $y=s-x$, integrate along the possible pairs that sum to $s$, and use the transformation $(x,y)\mapsto(x,s=x+y)$, whose Jacobian has absolute value 1. This gives the stated convolution integral.

Worked Example: A Discrete Sum With Several Routes to the Same Value

In a hypothetical data system, let $X$ have probabilities $P(X=0)=0.20$, $P(X=1)=0.50$, and $P(X=3)=0.30$. Independently, let $Y\sim\operatorname{Bernoulli}(0.40)$, so $P(Y=1)=0.40$ and $P(Y=0)=0.60$. Find the probability mass function of $S=X+Y$, $P(S\geq 3)$, and $E[S]$. This is a probability model, not an interest or mortality model; no interest rate or mortality assumption is involved.

Method: discrete convolution. The model gives $X$ and $Y$ explicitly and states that they are independent. List the possible pairs and their probabilities:

\begin{array}{c|c|c} X & Y & P(X,Y)\\ \hline 0&0&0.20(0.60)=0.12\\ 0&1&0.20(0.40)=0.08\\ 1&0&0.50(0.60)=0.30\\ 1&1&0.50(0.40)=0.20\\ 3&0&0.30(0.60)=0.18\\ 3&1&0.30(0.40)=0.12 \end{array}

Now combine pairs with equal sums. The sum 0 occurs only for $(0,0)$, and the sum 1 occurs only for $(0,1)$. The sum 2 occurs for $(1,1)$. The sum 3 occurs for $(3,0)$, and the sum 4 occurs for $(3,1)$. The remaining pair, $(1,0)$, gives sum 1, so it must be combined with $(0,1)$. Thus:

\begin{aligned} P(S=0)&=0.12,\\ P(S=1)&=0.08+0.30=0.38,\\ P(S=2)&=0.20,\\ P(S=3)&=0.18,\\ P(S=4)&=0.12. \end{aligned}

The requested tail probability is $P(S\geq3)=P(S=3)+P(S=4)=0.18+0.12=0.30$. The mean can be checked from the sum distribution:

$$ E[S]=0(0.12)+1(0.38)+2(0.20)+3(0.18)+4(0.12) =0+0.38+0.40+0.54+0.48=1.80. $$

As an independent check, $E[X]=0(0.20)+1(0.50)+3(0.30)=1.40$, while $E[Y]=0(0.60)+1(0.40)=0.40$; hence $E[X]+E[Y]=1.80$, as required. The probabilities also sum to $0.12+0.38+0.20+0.18+0.12=1.00$. The common mistake avoided here is assigning each sum to only one pair: sum 1 has two contributing pairs.

Continuous Sums: Supports Set the Bounds

For continuous convolution, the integrand alone is not enough: the limits must enforce both supports. For example, if $X$ can only lie in $[0,2]$ and $Y$ only in $[0,1]$, then at a fixed value $s$, a contributing $x$ must satisfy both $0\leq x\leq2$ and $0\leq s-x\leq1$. Equivalently, $x$ must lie in the intersection $[0,2]\cap[s-1,s]$. The intersection changes as $s$ moves across the support of the sum.

Worked Example: Convolving Two Uniform Distributions

Let $X\sim\operatorname{Uniform}(0,2)$ and $Y\sim\operatorname{Uniform}(0,1)$ be independent. Find the density of $S=X+Y$ and calculate $P(S\leq1.5)$. These are hypothetical measurement variables; no interest rate or mortality assumption is involved.

Method: continuous convolution with support-based limits. The densities are $f_X(x)=1/2$ on $0\leq x\leq2$, and $f_Y(y)=1$ on $0\leq y\leq1$, and each density is zero outside its stated interval. Therefore, the convolution integrand is $1/2$, and its integral is one-half the length of $[0,2]\cap[s-1,s]$.

The sum support is $0\leq s\leq3$. For $0<s<1$, the intersection is $[0,s]$, of length $s$. For $1\leq s<2$, it is $[s-1,s]$, of length 1. For $2\leq s<3$, it is $[s-1,2]$, of length $3-s$. Thus:

f_S(s)= \begin{cases} s/2,&0<s<1,\\ 1/2,&1\leq s<2,\\ (3-s)/2,&2\leq s<3,\\ 0,&\text{otherwise}. \end{cases}

To find the requested probability, integrate the density from the lower end of the support to 1.5. The integration crosses the point $s=1$, so split it there:

\begin{aligned} P(S\leq1.5) &=\int_0^1 \frac{s}{2}\,ds+\int_1^{1.5}\frac12\,ds\\ &=\left[\frac{s^2}{4}\right]_0^1+\left[\frac{s}{2}\right]_1^{1.5}\\ &=\frac14+\left(0.75-0.50\right)=0.50. \end{aligned}

Check the density by integrating over its full support: the first triangular section has area $1/4$, the constant section has area $(2-1)(1/2)=1/2$, and the last triangular section has area $1/4$. Their total is $1/4+1/2+1/4=1$. The common mistake avoided is using one set of bounds across the entire support; the overlap changes at $s=1$ and $s=2$.

Weighted Normal and Poisson Sums

Some sums can be handled directly once the assumptions are verified. If $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$ are independent, then $aX+bY$ is normal, with mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2+b^2\sigma_Y^2$. More generally, the variance of a weighted sum includes covariance: $ \operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y)$. Independence makes the covariance zero. For independent $X\sim\operatorname{Poisson}(\lambda_1)$ and $Y\sim\operatorname{Poisson}(\lambda_2)$, their sum has a Poisson distribution with parameter $\lambda_1+\lambda_2$.

Worked Example: A Weighted Sum of Independent Normal Variables

Suppose hypothetical measurement amounts, in thousands of dollars, satisfy $U\sim N(10,4)$ and $V\sim N(3,9)$, independently. Here the second argument of each normal distribution is the variance. Let $W=2U-V$. Find its distribution and $P(W>22)$. No interest rate or mortality assumption is part of this statistical model.

Method: the linear-combination normal result. Because $U$ and $V$ are independent normal variables, $W$ is normal. Its mean is $E[W]=2E[U]-E[V]=2(10)-3=17$. Its variance is

\operatorname{Var}(W)=2^2\operatorname{Var}(U)+(-1)^2\operatorname{Var}(V) =4(4)+1(9)=16+9=25.

Therefore, $W\sim N(17,25)$, with standard deviation $\sqrt{25}=5$. Standardize the cutoff 22:

P(W>22)=P\left(\frac{W-17}{5}>\frac{22-17}{5}\right) =P(Z>1)=1-\Phi(1)\approx 0.1587,

where $Z\sim N(0,1)$ and $\Phi$ is its cumulative distribution function. The answer is a probability, so it is reported to four decimal places. The check is that the cutoff is one standard deviation above the mean, giving the standard normal upper-tail probability at 1. The common mistake is to multiply the variance of $U$ by 2 rather than by $2^2$: that incorrect calculation gives $2(4)+9=17$, not the correct variance of 25.

Worked Example: A Tail Probability for a Poisson Sum

Two independent monitoring systems record event counts during the same fixed observation period. In a hypothetical model, $N\sim\operatorname{Poisson}(1.2)$ and $M\sim\operatorname{Poisson}(0.8)$. Find the probability that the combined count is at least 3. The parameters apply to this same period, and no interest rate or mortality assumption is involved.

Method: Poisson additivity. Since the counts are independent and both have Poisson distributions, $T=N+M\sim\operatorname{Poisson}(1.2+0.8)=\operatorname{Poisson}(2)$. Use the complement of counts 0, 1, and 2:

\begin{aligned} P(T\geq3) &=1-P(T=0)-P(T=1)-P(T=2)\\ &=1-e^{-2}\left(1+2+\frac{2^2}{2!}\right)\\ &=1-5e^{-2}\\ &\approx 1-5(0.1353353)\\ &\approx 0.3233. \end{aligned}

The probabilities in the complement are $P(T=0)=e^{-2}$, $P(T=1)=2e^{-2}$, and $P(T=2)=2^2e^{-2}/2!=2e^{-2}$, which sum to $5e^{-2}$. The final probability is rounded to four decimal places. The result is between 0 and 1, as required. The common mistake avoided is adding the two rates incorrectly or applying Poisson additivity without independence.

Checks That Protect the Calculation

Before accepting a sum distribution, make checks suited to the method. For a discrete convolution, add the probabilities and verify that no pair of outcomes was omitted. For a continuous convolution, verify that the density is zero outside the sum support and integrates to 1. For a weighted sum, check that each coefficient is applied to the mean once and to the variance squared. For a Poisson sum, check that the variables are independent and that the parameters refer to the same unit of observation.

These checks are especially useful when a question asks for a tail probability rather than a full distribution. A correct distribution can still lead to a wrong answer if the event is reversed or the complement includes the wrong endpoint. For integer-valued $T$, for example, $P(T\geq3)=1-P(T\leq2)$, so the subtracted values are exactly 0, 1, and 2.

Key takeaway: Identify the distribution, support, and dependence structure before choosing a sum method. Then check the result using total probability, support, parameters, or a direct mean calculation whenever possible.

Check Your Understanding

For each problem, identify a suitable method, carry out the calculation, and check the result against the stated assumptions.

Independent variables $X$ and $Y$ have $P(X=0)=0.70$, $P(X=2)=0.30$, $P(Y=1)=0.25$, and $P(Y=4)=0.75$. Find the probability mass function of $X+Y$. (Answer: $P(S=1)=0.175,\ P(S=3)=0.075,\ P(S=4)=0.525,\ P(S=6)=0.225$.)
Independent variables have distributions $X\sim\operatorname{Uniform}(0,1)$ and $Y\sim\operatorname{Uniform}(0,2)$. Use support-based convolution to find $P(X+Y\leq1)$. (Answer: $0.2500$.)
Independent variables $A\sim N(4,1)$ and $B\sim N(2,4)$ are combined as $R=3A+B$. Give the distribution of $R$ and find $P(R>14+\sqrt{13})$. (Answer: $R\sim N(14,13)$ and $P(R>14+\sqrt{13})\approx0.1587$.)
Independent counts have distributions $N\sim\operatorname{Poisson}(0.7)$ and $M\sim\operatorname{Poisson}(1.3)$. Find the probability that their total is exactly 2. (Answer: $P(N+M=2)=2e^{-2}\approx0.2707$.)
Independent variables $C\sim\operatorname{Binomial}(5,0.20)$ and $D\sim\operatorname{Binomial}(3,0.20)$ count successes in separate trials. Find $P(C+D\geq2)$. (Answer: $C+D\sim\operatorname{Binomial}(8,0.20)$, and $P(C+D\geq2)=1-0.8^8-8(0.2)(0.8^7)\approx0.4967$.)

What You'll Learn