Why MGF Identification Requires More Than Pattern Matching
A moment generating function can encode a distribution in a compact expression. Recognizing that expression is useful only if the parameters are valid and the function is genuinely an MGF on an interval around zero. A coefficient that looks like a probability but falls outside the probability range, or a domain that excludes zero, can invalidate an otherwise familiar-looking match.
For the standard distributions considered here, identification proceeds by matching the supplied expression to a known MGF, checking its parameters and domain, and then stating the distribution and its support. The uniqueness theorem for MGFs explains why the match determines the distribution: if two MGFs agree on an open interval containing zero, their probability distributions are the same.
Standard Forms and Their Checks
Use a consistent parameterization when matching formulas. In particular, the gamma MGF below uses the rate parameter \(\beta\), not the scale parameter. The support and domain are useful checks: a standard gamma variable is positive, while a normal variable can take any real value.
| Distribution | MGF \(M_X(t)\) | Parameter restrictions | Support and domain |
|---|---|---|---|
| Bernoulli | \(1-p+pe^t\) | \(0\leq p\leq 1\) | \(\{0,1\}\); all real \(t\) |
| Binomial | \((1-p+pe^t)^n\) | Positive integer \(n\); \(0\leq p\leq 1\) | \(\{0,1,\ldots,n\}\); all real \(t\) |
| Poisson | \(\exp(\lambda(e^t-1))\) | \(\lambda\geq 0\) | Nonnegative integers; all real \(t\) |
| Gamma (shape \(\alpha\), rate \(\beta\)) | \(\left(\frac{\beta}{\beta-t}\right)^\alpha\) | \(\alpha>0,\ \beta>0\) | Positive values; \(t<\beta\) |
| Normal | \(\exp\left(\mu t+\frac{\sigma^2t^2}{2}\right)\) | \(\mu\) real; \(\sigma^2>0\) | All real values; all real \(t\) |
The Bernoulli and binomial forms overlap when \(n=1\): a binomial variable with one trial is Bernoulli. A Poisson variable with \(\lambda=0\), or a Bernoulli variable with \(p=0\) or \(p=1\), is degenerate at one value. These boundary cases are valid when included by the parameter restrictions.
Why the Binomial Form Has This Shape
A short derivation shows why the binomial MGF is a power of a Bernoulli expression. For a binomial random variable \(X\) with parameters \(n\) and \(p\), the probability of \(X=k\) is \(\binom{n}{k}p^k(1-p)^{n-k}\), for \(k=0,1,\ldots,n\). Substituting these probabilities into the definition \(M_X(t)=E[e^{tX}]\) gives
The last equality is the binomial theorem. This derivation also explains why the power must be a positive integer for a standard binomial distribution and why \(p\) must be a probability. For all five distributions in the table, verify \(M_X(0)=1\). That check is necessary, though by itself it does not prove an expression is a valid MGF.
Worked Identifications
Worked Example: Recognizing a Bernoulli MGF
A hypothetical inspection indicator \(X\) has supplied MGF \(M_X(t)=0.18+0.82e^t\). Identify its distribution, parameter, support, and domain.
Method. Compare the expression with the Bernoulli form \(1-p+pe^t\). The coefficient of \(e^t\) is \(p\), so \(p=0.82\). The constant term must then be \(1-p=1-0.82=0.18\), which agrees with the supplied expression.
Thus \(X\sim\operatorname{Bernoulli}(0.82)\), with support \(\{0,1\}\). Its MGF is finite for all real \(t\), so it exists on an open interval containing zero. As a normalization check, \(M_X(0)=0.18+0.82=1\).
Common mistake avoided. The constant term is not \(p\); it is \(1-p\). Reading \(p=0.18\) would reverse the probabilities of 0 and 1 and would not match the coefficient of \(e^t\).
Worked Example: Checking a Binomial Match
A hypothetical system counts the number \(X\) of successful transmissions in a fixed batch. Its supplied MGF is \(M_X(t)=(0.65+0.35e^t)^6\). Identify the distribution and verify its parameters.
Method. Match the expression with \((1-p+pe^t)^n\). The exponent gives \(n=6\), and the coefficient of \(e^t\) gives \(p=0.35\). The remaining coefficient is \(0.65=1-0.35\). Both parameter checks pass: 6 is a positive integer and \(0.35\) lies between 0 and 1.
Therefore \(X\sim\operatorname{Binomial}(6,0.35)\), with support \(\{0,1,2,3,4,5,6\}\). Its MGF is finite for all real \(t\). It also passes the normalization check: \(M_X(0)=(0.65+0.35)^6=1^6=1\).
As an additional consistency check, the binomial mean is \(np=6(0.35)=2.1\), and its variance is \(np(1-p)=6(0.35)(0.65)=1.365\). These values are consistent with six trials and a success probability of 0.35.
Common mistake avoided. Do not interpret the exponent 6 as a probability parameter. It is the number of trials; the probability parameter is the coefficient \(0.35\) of \(e^t\).
Worked Example: Identifying a Poisson MGF
For a hypothetical count \(X\) of service requests during a fixed observation period, suppose \(M_X(t)=\exp(2.4(e^t-1))\). Identify the distribution, its parameter, support, and domain.
Method. Compare the exponent with the Poisson MGF \(\lambda(e^t-1)\). The multiplier is \(\lambda=2.4\), which is an admissible positive rate. Hence \(X\sim\operatorname{Poisson}(2.4)\).
A Poisson variable takes values \(0,1,2,\ldots\), and its MGF is finite for all real \(t\). Substituting \(t=0\) gives \(M_X(0)=\exp(2.4(1-1))=\exp(0)=1\). The mean and variance are both \(2.4\), providing another check on the identified rate.
Common mistake avoided. The parameter is not \(2.4e^t\) or the whole exponent as a function of \(t\). It is the constant multiplier \(\lambda=2.4\) of \(e^t-1\).
Worked Example: Distinguishing Gamma Rate from Scale
A hypothetical positive measurement \(X\) has MGF \(M_X(t)=\left(\frac{3}{3-t}\right)^4\). Identify its gamma parameters and state the domain.
Method. Match the expression with \(\left(\frac{\beta}{\beta-t}\right)^\alpha\). The exponent gives shape \(\alpha=4\), and the numerator and denominator give rate \(\beta=3\). Thus \(X\) has a gamma distribution with shape 4 and rate 3. Its equivalent scale is \(1/3\), not 3.
The support is positive, and the MGF is finite when \(t<3\). In particular, this domain contains an open interval around zero. At zero, \(M_X(0)=(3/3)^4=1\). The mean is \(\alpha/\beta=4/3\), and the variance is \(\alpha/\beta^2=4/9\).
Common mistake avoided. Calling 3 the scale would reverse the parameterization. With rate 3, the mean is \(4/3\); with scale 3 it would be 12, which does not match this MGF.
Worked Example: Reading a Normal MGF
A hypothetical error variable \(X\) has MGF \(M_X(t)=\exp(1.5t+1.125t^2)\). Identify the normal parameters and support.
Method. Compare the exponent with \(\mu t+\frac{\sigma^2t^2}{2}\). The coefficient of \(t\) gives \(\mu=1.5\). Equating the quadratic coefficients gives \(\sigma^2/2=1.125\), so \(\sigma^2=2(1.125)=2.25\). The standard deviation is \(\sigma=\sqrt{2.25}=1.5\).
Therefore \(X\sim N(1.5,2.25)\), where the second parameter is the variance. A normal variable has support over all real numbers, and its MGF is finite for all real \(t\). The normalization check gives \(M_X(0)=e^0=1\).
Common mistake avoided. The coefficient \(1.125\) is not the variance; it is half the variance. The variance is \(2.25\).
A Reliable Identification Procedure
When a supplied function is not already displayed in a standard form, rearrange it carefully and check each condition. A match based only on a visual resemblance is not enough. In particular, the existence-domain requirement matters for using uniqueness.
Confirm that the expression is defined at zero, that \(M_X(0)=1\), and that it is finite on an open interval containing zero.
Compare with the standard Bernoulli, binomial, Poisson, gamma, or normal MGF. Algebraic rewriting is acceptable, but the resulting parameters must remain admissible.
Verify integer restrictions, probability ranges, positive gamma shape and rate, and positive normal variance. Check that the supplied domain agrees with the candidate MGF near zero.
Give the parameterization explicitly, then name the support implied by the identified distribution. Apply MGF uniqueness to justify that this is the distribution.
A candidate expression that matches a formula only outside a neighborhood of zero does not meet the stated uniqueness condition. Similarly, an expression with an invalid parameter is not rescued by looking familiar. If the supplied information does not establish an MGF on an interval containing zero, do not claim that uniqueness has identified the distribution.
Common Exam Traps
- Reversing Bernoulli coefficients. In \(1-p+pe^t\), \(p\) multiplies \(e^t\), while \(1-p\) is the constant term.
- Accepting invalid binomial parameters. The exponent \(n\) must be a positive integer and \(p\) must lie in \([0,1]\).
- Confusing Poisson forms. The standard exponent is \(\lambda(e^t-1)\); the constant \(\lambda\) is the rate parameter.
- Mixing up gamma rate and scale. In \(\left(\frac{\beta}{\beta-t}\right)^\alpha\), \(\beta\) is the rate and the scale is \(1/\beta\). The domain restriction is \(t<\beta\).
- Forgetting the factor of one-half for a normal MGF. The coefficient of \(t^2\) is \(\sigma^2/2\), not \(\sigma^2\).
- Ignoring the neighborhood of zero. Matching an algebraic expression without checking where it is finite may not satisfy the condition required for MGF uniqueness.
Check Your Understanding
For each supplied MGF, identify the distribution and parameters, then check the support and domain.
- A hypothetical binary indicator has \(M_X(t)=0.2+0.8e^t\). Identify its Bernoulli parameter and state its support. (Answer: Bernoulli with \(p=0.8\); support \(\{0,1\}\); domain all real \(t\).)
- A batch count has \(M_X(t)=(0.7+0.3e^t)^8\). Identify its distribution and parameters. (Answer: Binomial with \(n=8,\ p=0.3\); support \(\{0,1,\ldots,8\}\); domain all real \(t\).)
- A hypothetical event count has \(M_X(t)=\exp(3(e^t-1))\). Identify its distribution and parameter. (Answer: Poisson with \(\lambda=3\); support \(0,1,2,\ldots\); domain all real \(t\).)
- A positive measurement has \(M_X(t)=\left(\frac{6}{6-t}\right)^2\). Give both the gamma rate and equivalent scale. (Answer: Gamma with shape \(2\), rate \(6\), and scale \(1/6\); support positive values; domain \(t<6\).)
- A measurement has \(M_X(t)=\exp(0.5t+1.125t^2)\). Identify its normal mean and variance, taking care to interpret the quadratic coefficient. (Answer: \(N(0.5,2.25)\); support all real values; domain all real \(t\).)