Common Errors in Sum Distributions

Why Sum Calculations Need Care

A sum can be easier to describe than either of its summands, but small errors in assumptions or bounds can produce a distribution that is impossible. A proposed density might be negative, assign probability outside the possible range, or integrate to something other than 1. A proposed variance might silently assume independence when the variables are dependent.

The main checks are straightforward: state the joint model, identify the support of the sum, use limits that satisfy both summands’ support restrictions, and check that the resulting probabilities and moments are consistent. These habits are useful whether the calculation uses convolution, conditioning, or moment-generating functions.

Definition: For random variables $X$ and $Y$, the support of $S=X+Y$ is the set of values that $X+Y$ can attain with positive probability (or, for continuous variables, values in the range where its density can be nonzero). If $X$ and $Y$ are independent with densities $f_X$ and $f_Y$, then $$ f_S(s)=\int_{-\infty}^{\infty} f_X(x)f_Y(s-x)\,dx. $$ The integrand is zero unless both $x$ is in the support of $X$ and $s-x$ is in the support of $Y$. For dependent variables, the corresponding formula uses their joint density $f_{X,Y}$: $$ f_S(s)=\int_{-\infty}^{\infty} f_{X,Y}(x,s-x)\,dx. $$

Error 1: Using Independence Without Justifying It

The product $f_X(x)f_Y(y)$ is the joint density only when $X$ and $Y$ are independent. Likewise, for discrete variables, $P(X=x,Y=y)=P(X=x)P(Y=y)$ requires independence. Knowing the marginal distributions alone does not determine the distribution of the sum when the variables may be dependent.

A related error is to calculate the variance of a sum as if the variables were independent. For variables with finite second moments, the correct identity is $$ \operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y). $$ The covariance term is zero when $X$ and $Y$ are independent, but independence is sufficient rather than necessary for zero covariance. If dependence is present, the covariance must be assessed rather than discarded.

Worked Example: A Shared Bernoulli Variable

Let $Z\sim\operatorname{Bernoulli}(0.5)$, and set $X=Z$ and $Y=Z$. Find the distribution, mean, and variance of $S=X+Y$. Is it correct to treat $S$ as $\operatorname{Binomial}(2,0.5)$? No interest or mortality assumptions apply; the probability model is specified by the shared Bernoulli variable.

Method and model. Since both summands equal the same random variable, use their joint dependence directly. If $Z=0$, then $S=0$; if $Z=1$, then $S=2$. Thus $$ P(S=0)=0.5,\qquad P(S=1)=0,\qquad P(S=2)=0.5. $$

Mean. From the distribution, $E[S]=0(0.5)+2(0.5)=1$. This agrees with $E[X]+E[Y]=0.5+0.5=1$.

Variance. The two possible values are each one unit from the mean, so $$ \operatorname{Var}(S)=(0-1)^2(0.5)+(2-1)^2(0.5)=1. $$ Equivalently, each Bernoulli variable has variance $0.5(0.5)=0.25$, and $\operatorname{Cov}(X,Y)=\operatorname{Var}(Z)=0.25$. Therefore, $$ \operatorname{Var}(S)=0.25+0.25+2(0.25)=1. $$

Common mistake and check. Treating the two variables as independent would give a $\operatorname{Binomial}(2,0.5)$ distribution, with positive probability $0.5$ at $S=1$ and variance $0.5$. But $S=1$ is impossible here, since $X=Y$ always. The proposed binomial model contradicts the joint construction, while the correct probabilities sum to $0.5+0+0.5=1$.

Error 2: Setting Convolution Bounds From Only One Support

In a continuous convolution, a valid integration value $x$ must meet two conditions at once: $x$ must be possible for $X$, and $s-x$ must be possible for $Y$. Ignoring either condition can create incorrect intervals or omit a change in the limits. The support of the sum itself also gives a fast check: if $X\in[a,b]$ and $Y\in[c,d]$, then $X+Y\in[a+c,b+d]$.

Worked Example: Finding Every Change in the Convolution Limits

Suppose $X\sim\operatorname{Uniform}(0,2)$ and $Y\sim\operatorname{Uniform}(1,4)$ are independent. Find the density of $S=X+Y$ and $P(2<S<4)$. No interest or mortality assumptions apply.

Method and support. The densities are $f_X(x)=1/2$ on $0\leq x\leq2$, and $f_Y(y)=1/3$ on $1\leq y\leq4$. The sum can range from $0+1=1$ to $2+4=6$. In the convolution, $0\leq x\leq2$ and $1\leq s-x\leq4$, so $s-4\leq x\leq s-1$. Combining these restrictions gives $$ \max(0,s-4)\leq x\leq\min(2,s-1). $$

Find the intervals. For $1<s<3$, the limits are $0$ and $s-1$. For $3\leq s\leq4$, they are $0$ and $2$. For $4<s<6$, they are $s-4$ and $2$. Therefore, $$ f_S(s)= \begin{cases} \dfrac{s-1}{6},&1<s<3,\\[4pt] \dfrac{1}{3},&3\leq s\leq4,\\[4pt] \dfrac{6-s}{6},&4<s<6,\\[4pt] 0,&\text{otherwise}. \end{cases} $$ The factors $1/6$ in the rising and falling sections are the product of the two densities, $1/2$ and $1/3$.

Calculate the requested probability. Split the interval at the density’s change point $s=3$: $$ P(2<S<4)=\int_2^3\frac{s-1}{6}\,ds+\int_3^4\frac{1}{3}\,ds =\frac{1}{4}+\frac{1}{3} =\frac{7}{12} \approx 0.5833. $$ The probability is reported to 4 decimals.

Common mistake and check. A frequent error is to use the decreasing section as soon as $s>3$. That misses the plateau: for $3<s<4$, the correct limits remain $0$ to $2$, so the density is $2/6=1/3$. Check the full density: the increasing section has area $1/3$, the plateau has area $1/3$, and the decreasing section has area $1/3$. Their total is 1, as required.

A Short Derivation: Why the Convolution Limits Matter

For independent continuous variables, the joint density is $f_X(x)f_Y(y)$. To find the density of $S=X+Y$, fix a value $s$, write $y=s-x$, and add the joint-density contributions from all pairs $(x,y)$ on the line $x+y=s$. Integrating along that line gives $$ f_S(s)=\int_{-\infty}^{\infty}f_X(x)f_Y(s-x)\,dx. $$ This integral includes only feasible pairs because a density is zero outside its support. When the densities have bounded supports, it is usually clearer to solve both support inequalities explicitly, as in the preceding example, before integrating.

The formula also explains two essential checks. First, the possible values of $s$ are restricted by the summands’ supports. Second, integrating the density over all $s$ must give 1: every possible pair $(X,Y)$ contributes to exactly one sum value $X+Y$. A result that fails either check signals an error in the setup or calculation.

Error 3: Applying a Familiar Distribution Result Outside Its Conditions

Some sum results require more than a distribution name. For example, the sum of independent Bernoulli variables with a common success probability $p$ is binomial. With different success probabilities, the sum is still a valid discrete random variable, but it need not have a binomial distribution. The same principle applies to other special distribution results: verify their assumptions before using them.

Worked Example: Independent Bernoulli Variables With Unequal Probabilities

Let $X\sim\operatorname{Bernoulli}(0.2)$ and $Y\sim\operatorname{Bernoulli}(0.6)$ be independent. Find the probability mass function, mean, and variance of $S=X+Y$. No interest or mortality assumptions apply.

Method and support. Each variable can be 0 or 1, so $S\in\{0,1,2\}$. Independence gives $$ P(S=0)=P(X=0)P(Y=0)=0.8(0.4)=0.32, $$ $$ P(S=1)=P(X=0,Y=1)+P(X=1,Y=0) =0.8(0.6)+0.2(0.4)=0.48+0.08=0.56, $$ and $$ P(S=2)=P(X=1)P(Y=1)=0.2(0.6)=0.12. $$

Check the distribution. All probabilities are nonnegative, and $0.32+0.56+0.12=1$. Thus the mass function is complete and consistent with the support.

Mean and variance. From the sum of expectations, $E[S]=0.2+0.6=0.8$. Independence means the covariance is zero, so $$ \operatorname{Var}(S)=0.2(0.8)+0.6(0.4)=0.16+0.24=0.40. $$ The mass function gives the same variance: $E[S^2]=0^2(0.32)+1^2(0.56)+2^2(0.12)=0.56+0.48=1.04$, and $1.04-(0.8)^2=0.40$.

Common mistake and check. Using $\operatorname{Binomial}(2,0.4)$, with $0.4$ obtained by averaging the success probabilities, would give $P(S=0)=0.36$, $P(S=1)=0.48$, and $P(S=2)=0.16$. Those values do not match the probabilities obtained from the actual independent model. The common-$p$ condition for a binomial sum is not met.

Error 4: Confusing Distribution Parameters

A formula can be used correctly but still give a wrong result if the parameter convention is misread. In particular, an exponential or gamma distribution may be parameterized by a rate or by a scale. For an exponential distribution with rate $\lambda$, the density is $\lambda e^{-\lambda x}$ for $x>0$, with mean $1/\lambda$ and variance $1/\lambda^2$. Its equivalent scale is $1/\lambda$, not $\lambda$.

Worked Example: Summing Two Exponentials With a Rate Parameter

Suppose $X$ and $Y$ are independent exponential variables, each with rate $2$. Find the density, mean, and variance of $S=X+Y$, and identify its gamma parameters. No interest or mortality assumptions apply.

Method. Each density is $2e^{-2x}$ for $x>0$, and the sum has support $s>0$. For a fixed $s>0$, both summands are positive when $0<x<s$. Thus $$ f_S(s)=\int_0^s 2e^{-2x}\,2e^{-2(s-x)}\,dx =4e^{-2s}\int_0^s 1\,dx =4s e^{-2s},\qquad s>0. $$ The density is zero for $s\leq0$.

Identify the distribution and moments. This is a gamma distribution with shape $2$ and rate $2$. Equivalently, it has shape $2$ and scale $1/2$. By additivity of expectations and variances for independent variables, $$ E[S]=\frac{1}{2}+\frac{1}{2}=1,\qquad \operatorname{Var}(S)=\frac{1}{2^2}+\frac{1}{2^2}=\frac{1}{2}. $$

Common mistake and check. Reading the stated rate $2$ as a scale of $2$ would instead imply an exponential mean of $2$, contradicting the rate-parameter mean $1/2$. For the correct density, the total probability is $$ \int_0^\infty 4s e^{-2s}\,ds=1. $$ The positive support, unit mean, variance $1/2$, and total probability of 1 are consistent with the rate-2 model.

A Reliable Checklist for Sum Problems

Before accepting a calculated sum distribution, work through these checks. They are especially useful when a result is piecewise or when a named distribution seems to offer a shortcut.

State the joint assumption.
Write whether the variables are independent or specify the dependence. Do not use a product of marginal densities or probabilities without independence.

Find the sum’s support.
Combine the possible values of both summands. Reject any proposed density or mass outside that range.

Impose both support restrictions.
In a convolution, require both $x$ and $s-x$ to be possible. Look for every point where the limits change.

Check the parameter convention.
For a named distribution, verify whether each parameter is a rate, scale, probability, or variance.

Validate the result.
Check nonnegative probabilities, total mass or area 1, and consistency between moments computed by different valid methods.

For the variance, keep the dependence visible until it has been justified away. In general, $$ \operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y), $$ and for constants $a$ and $b$, $$ \operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y). $$ The covariance term vanishes under independence, but the marginal distributions by themselves do not establish independence.

Key takeaway: A correct calculation of a sum begins with the joint model, not just the marginal distributions. Then check the support, convolution limits, parameter convention, and covariance term. These checks are aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, check the support and assumptions as well as the requested numerical result.

Independent variables $X\sim\operatorname{Bernoulli}(0.3)$ and $Y\sim\operatorname{Bernoulli}(0.5)$ are summed. Find $P(X+Y=1)$, $E[X+Y]$, and $\operatorname{Var}(X+Y)$. (Answer: $0.5000,\ 0.8,\ 0.46$)
Let $Z\sim\operatorname{Bernoulli}(0.4)$, and define $X=Z$, $Y=1-Z$. Give the distribution and variance of $X+Y$; explain why an independence calculation would be inappropriate. (Answer: $P(X+Y=1)=1$, variance $0$; the variables are dependent and their sum is constant.)
Independent variables $X\sim\operatorname{Uniform}(0,1)$ and $Y\sim\operatorname{Uniform}(0,2)$ are summed. Find the support of $S=X+Y$ and $P(S<1)$. (Answer: support $[0,3]$, $P(S<1)=0.25$)
Let $X$ and $Y$ be independent exponentials, each with rate $3$. Find the mean and variance of their sum, and the probability that the sum exceeds 1. (Answer: mean $2/3$, variance $2/9$, $P(X+Y>1)=4e^{-3}\approx0.1991$)
Let $X\sim N(2,4)$ and $Y\sim N(-1,9)$ be independent, with the second argument denoting variance. Find the distribution of $T=2X-Y$, including its mean and variance. (Answer: $T\sim N(5,25)$)

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