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Exam P · Sums of Independent Variables · Tutorial 820 of 915

Identifying Sum Distributions from MGFs

Use the product of independent variables’ MGFs to recognize a sum’s distribution, verify that the match is valid, and avoid common parameter and probability errors.

Advanced 11 min read

What You'll Learn

  • Multiply MGFs for independent summands and determine where the product is valid.
  • Match a product MGF to a known distribution’s standard form.
  • Identify a normal sum by reading its mean and variance from the exponent.
  • Use uniqueness of MGFs to justify a distributional identification without convolution.
  • Check probabilities from an identified sum using the correct variance and unrounded CDF values.
  • Detect an impossible proposed MGF by checking its implied variance.

Why Recognizing a Sum MGF Helps

A sum of independent random variables can be studied by finding its distribution directly or by combining the distributions of its summands. A convolution may require several cases or an integral. When the summands have known moment generating functions, multiplying those MGFs can reveal the distribution of the sum without carrying out that convolution.

The key is not just to multiply. The product must be matched carefully to the standard MGF of a candidate distribution, on a valid interval around zero. That match, together with uniqueness of MGFs, establishes the distribution of the sum. This method is aligned with the SOA Exam P (Probability) syllabus.

Definition: The moment generating function (MGF) of a random variable \(X\) is \(M_X(t)=E[e^{tX}]\), for values of \(t\) for which the expectation is finite. If \(X\) and \(Y\) are independent and their MGFs exist in a neighborhood of zero, then the MGF of \(S=X+Y\) is \(M_S(t)=M_X(t)M_Y(t)\) wherever both factors are defined. An MGF that exists in a neighborhood of zero uniquely determines the distribution.

A Reliable Identification Method

For a sum \(S=X_1+\cdots+X_n\) of independent random variables, begin with the MGF product rule established in the earlier material on the MGF product rule for independent sums. Multiply the MGFs, simplify the result, and compare it with a known MGF. For example, the MGF of a normal random variable \(N(\mu,\sigma^2)\), where the second parameter is the variance, is

$$ M_X(t)=\exp\left(\mu t+\frac{\sigma^2t^2}{2}\right). $$

Thus, if a product simplifies to \(\exp(at+bt^2)\), it matches a normal MGF with mean \(a\) and variance \(2b\), provided \(2b>0\). The coefficient of \(t^2\) is half the variance. If \(b=0\), the expression is the MGF of the constant \(a\); if \(b<0\), it cannot be the MGF of a normal random variable with positive variance.

1
Check the assumptions.
Confirm the summands are independent and their MGFs exist together on an open interval containing zero.
2
Multiply the MGFs.
Use independence to express the sum’s MGF as the product of the individual MGFs.
3
Match the standard form.
Rearrange the product and compare its parameters with a candidate distribution’s MGF.
4
Verify the identification.
Check that the parameters are valid and the MGFs agree near zero. Then invoke uniqueness.

The product rule follows directly from independence. For \(S=X+Y\),

$$ M_S(t)=E[e^{t(X+Y)}] =E[e^{tX}e^{tY}] =E[e^{tX}]E[e^{tY}] =M_X(t)M_Y(t). $$

The middle equality separating the expectation is exactly where independence is used. If the variables are dependent, the product rule need not hold. Once the product agrees with a candidate distribution’s MGF on an open interval around zero, uniqueness says that the sum and the candidate have the same distribution. Equality at a single value of \(t\), or a formal match outside the domain where the MGFs exist, is not sufficient.

Worked Example: Matching a Product to a Normal MGF

Independent variables \(X\) and \(Y\) have MGFs \(M_X(t)=e^{4t+3t^2}\) and \(M_Y(t)=e^{-t+2t^2}\). Identify the distribution of \(S=X+Y\), then find \(P(3-\sqrt{10}<S<3+\sqrt{10})\).

Method. Multiply the MGFs, match the result to the normal MGF, and standardize the requested interval. This avoids treating the coefficient of \(t^2\) as the variance.

By independence,

$$ M_S(t)=e^{4t+3t^2}e^{-t+2t^2} =e^{(4-1)t+(3+2)t^2} =e^{3t+5t^2}. $$

The normal MGF has exponent \(\mu t+\sigma^2t^2/2\). Matching coefficients gives \(\mu=3\) and \(\sigma^2/2=5\), so \(\sigma^2=10\). Therefore \(S\sim N(3,10)\). The variance is 10, not 5.

Standardize the two endpoints using \(\sigma=\sqrt{10}\):

$$ P(3-\sqrt{10}<S<3+\sqrt{10}) =P(-1<Z<1) =\Phi(1)-\Phi(-1) =0.8413447-0.1586553 =0.6826894. $$

To four decimal places, the probability is 0.6827. The result is checked by symmetry: the interval is centered at the mean and extends one standard deviation on each side.

Matching Parameters and Calculating Probabilities

Sometimes the MGF product contains an unknown parameter, or a probability calculation requires careful attention to both the mean and variance. Keep the identification separate from the probability step: first establish the distribution, then standardize using its parameters.

Worked Example: A Normal Sum and a CDF Difference

Three independent variables have MGFs \(e^{2t+t^2}\), \(e^{-3t+4t^2}\), and \(e^{t^2/2}\). Let their sum be \(S\). Identify its distribution and calculate \(P(-1<S<5)\).

Method. Add the linear and quadratic terms in the exponents, match to the normal MGF, and use the standard normal CDF. This avoids confusing a CDF value with the probability between two endpoints.

The product is

$$ M_S(t)=\exp\left((2-3+0)t+ \left(1+4+\frac12\right)t^2\right) =\exp\left(-t+\frac{11}{2}t^2\right). $$

Matching with \(\exp(\mu t+\sigma^2t^2/2)\) gives \(\mu=-1\) and \(\sigma^2/2=11/2\), hence \(\sigma^2=11\). Thus \(S\sim N(-1,11)\).

Standardize each endpoint:

$$ P(-1<S<5) =P\left(0<\frac{S-(-1)}{\sqrt{11}}<\frac{6}{\sqrt{11}}\right) =\Phi\left(\frac{6}{\sqrt{11}}\right)-\Phi(0). $$

Here \(6/\sqrt{11}\approx1.8091\), \(\Phi(6/\sqrt{11})\approx0.9648\), and \(\Phi(0)=0.5000\). Using unrounded CDF values, the probability is approximately \(0.4648\) to four decimal places. The lower endpoint is the mean, so a result less than one-half is consistent with the upper endpoint being less than two standard deviations above the mean.

Worked Example: Inferring an Unknown Mean Before Matching

Independent variables \(X\) and \(Y\) have MGFs \(M_X(t)=e^{2t+t^2}\) and \(M_Y(t)=e^{bt+3t^2}\), where \(b\) is unknown. It is also known that \(E[X+Y]=-1\). Find \(b\), identify the distribution of \(S=X+Y\), and calculate \(P(-1<S<-1+\sqrt{8})\).

Method. Use the linear coefficient of the product MGF, which is the mean, to determine \(b\); then match both coefficients to the normal MGF. This avoids assuming the unknown parameter is zero.

The sum MGF is

$$ M_S(t)=e^{(2+b)t+4t^2}. $$

For this normal-form MGF, the mean is \(2+b\). The given mean condition therefore yields \(2+b=-1\), so \(b=-3\). The quadratic coefficient is \(4=\sigma^2/2\), giving \(\sigma^2=8\). Consequently \(S\sim N(-1,8)\).

Since the interval begins at the mean and ends one standard deviation above it,

$$ P(-1<S<-1+\sqrt{8}) =P(0<Z<1) =\Phi(1)-\Phi(0) =0.8413447-0.5000000 =0.3413447. $$

To four decimal places, the probability is 0.3413. The check is consistent with a one-standard-deviation interval on only one side of a symmetric distribution.

When the Match Is Valid—and When It Is Not

A familiar-looking formula is useful only if its parameters and domain make sense. For a normal candidate, the coefficient of \(t^2\) must be nonnegative, because it equals half the variance. More generally, a proposed MGF must satisfy basic properties such as \(M_X(0)=1\), and, when the MGF exists around zero, its first two derivatives at zero give \(E[X]\) and \(E[X^2]\). The implied variance \(E[X^2]-(E[X])^2\) cannot be negative.

The independence assumption also deserves an explicit check. Knowing the marginal MGFs of \(X\) and \(Y\) does not identify the MGF of \(X+Y\) by multiplication unless \(X\) and \(Y\) are independent. In addition, a product must be valid on a common interval around zero: if one factor exists only for \(t\) in one interval and the other only in a disjoint interval, the uniqueness argument cannot be applied through that product.

  • Wrong: From \(e^{3t+5t^2}\), report a normal variance of 5 because 5 multiplies \(t^2\).
  • Correction: Compare with \(e^{\mu t+\sigma^2t^2/2}\). The variance is twice the coefficient, so it is 10.
  • Wrong: Identify a sum by multiplying the component MGFs even though the variables may be dependent.
  • Correction: Verify independence before using the product rule; otherwise the joint distribution may be needed.
  • Wrong: Treat \(\Phi(z)\) as the probability of an interval ending at \(z\).
  • Correction: Subtract the lower endpoint’s CDF value: \(P(a<Z<b)=\Phi(b)-\Phi(a)\).
Key takeaway: An independent-sum MGF identifies a distribution when the product is valid near zero and matches that distribution’s MGF there. Match every parameter, especially the factor of one-half in the normal quadratic term, and use uniqueness only after checking the assumptions.

Check Your Understanding

For each problem, use the product MGF rule when independence is stated, and match parameters before calculating any probability.

  1. Independent \(U\) and \(V\) have MGFs \(e^{t+2t^2}\) and \(e^{3t+t^2/2}\). Identify the distribution of \(U+V\). (Answer: \(N(4,5)\).)
  2. Independent \(A\) and \(B\) have MGFs \(e^{-2t+3t^2}\) and \(e^{t+2t^2}\). Identify the distribution of \(A+B\). (Answer: \(N(-1,10)\).)
  3. Independent \(R\) and \(T\) have MGFs \(e^{t+2t^2}\) and \(e^{-4t+3t^2}\). Identify the distribution of \(R+T\), then find \(P(R+T<-3+\sqrt{10})\). (Answer: \(N(-3,10)\), and \(0.8413\).)
  4. Three independent variables have MGFs \(e^{2t+t^2}\), \(e^{-3t+4t^2}\), and \(e^{t^2/2}\). Identify their sum and calculate the probability that it lies between \(-1\) and \(5\). (Answer: \(N(-1,11)\), and \(0.4648\).)
  5. A proposed sum MGF is \(e^{2t-t^2}\) on a neighborhood of zero. Can it be the MGF of a random variable? Check its implied mean and variance. (Answer: No; the implied mean is \(2\) and variance is \(-2\), which is impossible.)

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