Shifts and Scales Before Summing

Why Transform Each Summand First?

A sum problem may describe variables in a form that hides their familiar distributions. For example, a summand might be twice an exponential variable, a negative multiple of a discrete count, or a shifted uniform variable. Before choosing a sum method, identify what each transformed summand can actually take and how its distribution has changed. That step can reveal a standard distributional closure, or provide the correct inputs for convolution.

Throughout, assume the original summands are independent unless stated otherwise. Applying a deterministic function separately to independent variables preserves independence. Thus, if \(X_1,\ldots,X_n\) are independent and \(Y_j=a_jX_j+b_j\), then the \(Y_j\) are also independent.

Definition: An affine transformation of a random variable \(X\) is \(Y=aX+b\), where \(a\) and \(b\) are constants. The constant \(b\) shifts the values, while \(a\) scales them and, if negative, reverses their order. For \(a\ne 0\), the support of \(Y\) is the set of values \(ax+b\) as \(x\) ranges over the support of \(X\).

A shift changes location but not the spacing between possible values. A scale changes that spacing; a negative scale also reverses the direction. These support changes are not cosmetic: they determine which values can contribute to a sum and therefore which limits belong in a convolution.

What a Shift and Scale Do to a Distribution

For \(Y=aX+b\) with \(a\ne 0\), probabilities can be found by translating the event back to \(X\). In particular, when \(a>0\), \(F_Y(y)=F_X((y-b)/a)\). When \(a<0\), the inequality reverses, so \(F_Y(y)=P(X\ge (y-b)/a)\); if \(X\) is continuous this is \(1-F_X((y-b)/a)\). For a continuous \(X\) with density \(f_X\), the density in either case is

f_Y(y)=\frac{1}{|a|}f_X\!\left(\frac{y-b}{a}\right).

The absolute value accounts for the change in interval length even when the transformation reverses direction. For a discrete variable, each value \(x\) with probability \(P(X=x)\) maps to \(ax+b\) with that same probability; for \(a\ne0\), distinct values remain distinct.

The transformation rules for the first two moments follow directly from linearity of expectation and the definition of variance:

E[aX+b]=aE[X]+b,\qquad \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X).

The square in the variance rule matters: multiplying outcomes by \(a\) multiplies every deviation from the mean by \(a\), and variance averages the squared deviations. A shift cancels when subtracting the mean.

Transforming Before Combining Independent Variables

The same idea applies to moment-generating functions. When the relevant moment-generating function exists, the definition \(M_X(t)=E[e^{tX}]\) gives the affine transformation rule below. For independent summands, their transformed moment-generating functions multiply. This can identify a sum’s distribution when the product matches a known form.

M_{aX+b}(t)=e^{bt}M_X(at), \qquad M_{\sum_{j=1}^n(a_jX_j+b_j)}(t) =\prod_{j=1}^n e^{b_jt}M_{X_j}(a_jt).

Transform each variable.
Find the support and distribution of each \(a_jX_j+b_j\), including any changed parameters.

Check independence and the sum method.
Independence permits a product of moment-generating functions or the usual convolution formula. Then check whether the transformed summands meet the conditions for a standard distributional closure.

Combine and check.
Apply the closure rule or convolve the transformed distributions. Verify that the resulting support and, when useful, moments are consistent with the original sum.

For example, if \(X\) has a gamma distribution with shape \(\alpha\) and rate \(\lambda\), then \(cX\), for \(c>0\), has a gamma distribution with shape \(\alpha\) and rate \(\lambda/c\). This follows by substituting \(x=y/c\) into the density and including the density factor \(1/c\). A positive scale changes the rate in the denominator of the exponential term; it does not multiply the rate. A fixed shift adds a location shift and generally means the result is no longer an ordinary gamma distribution on the nonnegative support.

Worked Example: Scaling Exponentials to Match Gamma Rates

Let \(U\) and \(V\) be independent exponential random variables with rates \(4\) and \(6\), respectively. Find the distribution of \(S=5+2U+3V\), and calculate \(P(S\le 7)\).

Method. Transform each exponential first, then use gamma closure for independent exponentials with the same rate. The common mistake avoided here is multiplying an exponential rate by the scale factor.

An exponential variable with rate \(\lambda\) has mean \(1/\lambda\) and density \(\lambda e^{-\lambda x}\) for \(x\ge0\). For \(c>0\), \(P(cU>y)=P(U>y/c)=e^{-\lambda y/c}\) for \(y\ge0\), so \(cU\) is exponential with rate \(\lambda/c\). Therefore, \(2U\) has rate \(4/2=2\), and \(3V\) has rate \(6/3=2\).

The transformed variables are independent exponentials with common rate \(2\). Their sum has a gamma distribution with shape \(2\) and rate \(2\). Thus \(2U+3V\sim\operatorname{Gamma}(2,2)\), using the shape-rate parameterization, and \(S\) is this gamma variable shifted by \(5\).

The event \(S\le7\) is equivalent to \(2U+3V\le2\). The gamma density for shape \(2\) and rate \(2\) is \(4t e^{-2t}\) for \(t\ge0\), and its cumulative distribution function is \(1-e^{-2t}(1+2t)\). Hence

P(S\le7) =1-e^{-2(2)}(1+2(2)) =1-5e^{-4} \approx 0.9084.

The support of \(S\) is \([5,\infty)\), so the cutoff \(7\) is in the support. The calculated probability is between \(0\) and \(1\), as required.

When Transformed Variables Need Convolution

Not every transformed sum matches a named distribution. For two independent continuous variables with densities \(f_A\) and \(f_B\), the sum \(T=A+B\) has density \(f_T(t)=\int_{-\infty}^{\infty} f_A(a)f_B(t-a)\,da\), with the effective integration limits determined by the supports. For discrete variables, the corresponding probabilities are summed over support values that add to the target. Transform the variables first; then impose the resulting supports in the convolution.

Worked Example: Convolving Two Scaled Uniform Variables

Let \(U\sim\operatorname{Unif}(0,1)\) and \(V\sim\operatorname{Unif}(0,2)\) be independent. Set \(S=3+2U+\tfrac12 V\). Find the density of \(S\) and calculate \(P(S\le4.5)\).

Method. Transform both uniforms, identify their supports and densities, and convolve. The trap avoided is convolving the original uniforms without adjusting their supports and density heights.

Let \(A=2U\) and \(B=\tfrac12V\). Then \(A\) is uniform on \([0,2]\), with density \(1/2\), while \(B\) is uniform on \([0,1]\), with density \(1\). The sum \(T=A+B\) has support \([0,3]\). For a given \(t\), the convolution integrates the product of the two densities over values \(a\) satisfying both \(0\le a\le2\) and \(0\le t-a\le1\). Equivalently, \(\max(0,t-1)\le a\le\min(2,t)\). The integrand is \(1/2\) wherever those support intervals overlap. This gives

f_T(t)= \begin{cases} t/2, & 0<t<1,\\ 1/2, & 1\le t<2,\\ (3-t)/2, & 2\le t<3,\\ 0, & \text{otherwise}. \end{cases}

Since \(S=3+T\), its density is \(f_S(s)=f_T(s-3)\): the same shape shifted to support \([3,6]\). The probability asked for is \(P(T\le1.5)\). The area from \(0\) to \(1\) is \(\int_0^1 t/2\,dt=1/4\), and the area from \(1\) to \(1.5\) is \(\int_1^{1.5}1/2\,dt=1/4\). Therefore \(P(S\le4.5)=1/2=0.5000\). As a check, the full density has total area \(1/4+1/2+1/4=1\).

Shifts and Scales for Normal Sums

A normal variable remains normal after an affine transformation. More specifically, if \(X\sim N(\mu,\sigma^2)\), then \(aX+b\sim N(a\mu+b,a^2\sigma^2)\). If the transformed variables are independent normal variables, their sum is also normal, with mean equal to the sum of their means and variance equal to the sum of their variances. The variance contributions use squared coefficients; signs affect the mean but not the variance.

Worked Example: A Linear Combination of Independent Normals

Suppose \(X\sim N(4,9)\) and \(Y\sim N(2,1)\) are independent. For \(S=2X-3Y+1\), find its distribution and calculate \(P(S>3+3\sqrt{5})\).

Method. Transform the two independent normal variables and add their means and variances. The negative coefficient on \(Y\) reverses its values and changes its contribution to the mean, but it does not make its variance negative.

The transformed means are \(E[2X]=2(4)=8\) and \(E[-3Y]=-3(2)=-6\). Including the constant, \(E[S]=8-6+1=3\). The transformed variances are \(\operatorname{Var}(2X)=2^2(9)=36\) and \(\operatorname{Var}(-3Y)=(-3)^2(1)=9\). Independence gives \(\operatorname{Var}(S)=36+9=45\). Thus \(S\sim N(3,45)\).

Because \(\sqrt{45}=3\sqrt5\), the requested cutoff is one standard deviation above the mean. Standardizing,

P(S>3+3\sqrt5) =P\!\left(\frac{S-3}{\sqrt{45}}>1\right) =P(Z>1) \approx 0.1587,

where \(Z\sim N(0,1)\). The result is a reasonable upper-tail probability for a point one standard deviation above the mean.

Common Mistake: Scaling a Rate or Variance Incorrectly

A frequent error is to multiply a distribution’s rate by the scale factor, or to multiply a variance by the signed coefficient instead of its square. For the exponential variable \(U\) with rate \(4\) in the first example, an incorrect calculation might claim \(2U\) has rate \(8\). That would imply \(P(2U>1)=e^{-8}\). But the event itself gives

P(2U>1)=P(U>1/2)=e^{-4(1/2)}=e^{-2}.

Therefore, the correct rate is \(2\), not \(8\). For a scaled variable \(aX+b\), the correct variance is \(a^2\operatorname{Var}(X)\), regardless of whether \(a\) is positive or negative. Finally, shifting each summand contributes to the total location: if \(S=\sum_j(a_jX_j+b_j)\), then every \(b_j\) must be included in its mean and in its support.

Key takeaway: Before summing, transform every summand separately. Check its support, distribution parameters, and scale-adjusted density or probabilities. Only then apply an appropriate closure rule or convolution, and verify that the resulting support and moments make sense.

Check Your Understanding

For each independent-sum problem, transform the variables before combining them. Answers are included so you can check your calculations.

Let \(X\) and \(Y\) be independent exponential variables with rates \(3\) and \(8\). For \(S=5+2X+4Y\), find the support, mean, and variance. (Answer: support \([5,\infty)\), \(E[S]=37/6\), \(\operatorname{Var}(S)=25/36\).)
Let \(A\sim\operatorname{Unif}(-1,2)\) and \(B\sim\operatorname{Unif}(0,1)\) be independent. For \(S=3-2A+2B\), find the lower and upper endpoints of its support. (Answer: support \([-1,7]\).)
Let \(X\sim N(1,4)\) and \(Y\sim N(3,9)\) be independent. For \(S=4X-Y+2\), find its distribution and \(P(S>3+\sqrt{73})\). (Answer: \(S\sim N(3,73)\); probability \(0.1587\).)
Let \(X\sim\operatorname{Gamma}(2,3)\) and \(Y\sim\operatorname{Gamma}(3,5)\) be independent, with shape-rate parameterization. For \(S=4+2X+3Y\), calculate its mean and variance, and state whether the two transformed gamma variables have a common rate. (Answer: mean \(107/15\), variance \(443/225\); the rates are \(3/2\) and \(5/3\), so they are not equal.)
Let \(N\sim\operatorname{Poisson}(2)\) and define \(S=7-3N\). Find the support of \(S\) and \(P(S=1)\). (Answer: support \(\{7,4,1,-2,\ldots\}\); \(P(S=1)=P(N=2)=2e^{-2}\approx0.2707\).)

What You'll Learn