Likelihoods Up to Parameter-Free Multipliers

Why Likelihoods Are Often Written in Simplified Form

A likelihood can contain factorials, powers of known exposures, and exponential terms involving observed data. When the goal is to find a maximum likelihood estimate, some of these terms can be removed to make the maximization easier. The key question is not whether a term looks complicated or “constant,” but whether it changes as any unknown parameter changes.

This distinction matters in actuarial applications, where likelihoods may be built from claim counts, policy-level observations, or measurements with a specified probability model. A simplified expression can make the maximizing parameter easier to see, but an invalid omission can change the optimization problem itself. The ideas here are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $L(\theta;x)$ be a likelihood for observed data $x$, where $\theta$ denotes all unknown parameters being estimated. A likelihood expression is proportional to $L$ with respect to $\theta$ if it has the form $$ L^*(\theta;x)=c(x)L(\theta;x), $$ where $c(x)$ is positive and does not depend on any unknown parameter in $\theta$. Then $L^*$ and $L$ have the same maximizing parameter value or values.

The factor $c(x)$ may depend on the observed data, the sample size, or a fixed design value such as a known exposure. Those quantities are held fixed while the likelihood is maximized. It must not depend on any unknown parameter, including a nuisance parameter that is being estimated jointly or profiled.

The Safe-Omission Rule

For a fixed observed data set, removing a positive parameter-free factor rescales every likelihood value by the same amount. It therefore preserves their ordering: if one parameter value gave a larger likelihood before the rescaling, it still gives a larger likelihood afterward.

L(\theta_1;x)>L(\theta_2;x) \quad\Longleftrightarrow\quad c(x)L(\theta_1;x)>c(x)L(\theta_2;x), \qquad c(x)>0.

Consequently, the set of values of $\theta$ that maximize the likelihood is unchanged. The same reasoning applies to the log likelihood: multiplying by a positive factor adds the fixed quantity $\log c(x)$ to the log likelihood, which does not change its maximizer.

\ell^*(\theta;x) =\log L^*(\theta;x) =\log L(\theta;x)+\log c(x).

This is a justification for simplifying an expression before maximizing it; it is not permission to remove any term that happens not to contain the data in an obvious way. A factor such as $\exp(-\theta)$ depends on $\theta$ and cannot be removed. Nor can a factor be dropped merely because it does not contain one of several unknown parameters: it must be independent of all unknown parameters in the optimization.

Write the likelihood from the model.
Include the probability mass or density contributions for the observed data and any fixed design information.

Mark every unknown parameter.
Include all parameters being estimated, not only the one you plan to solve for first.

Separate parameter-free factors.
A positive factor may be omitted only if its value is unchanged for every candidate value of every unknown parameter.

Maximize and check the model.
Use the simplified expression to find the maximizer, then verify that the omitted terms really were parameter-free.

Worked Examples

Worked Example: Independent Binomial Counts

A hypothetical insurer divides a study portfolio into three fixed groups. In the groups, respectively, $3$ of $8$, $4$ of $10$, and $1$ of $6$ policies have a specified claim feature. Assume the counts are independent and each group count has a binomial distribution with its stated size and a common unknown probability $p$, where $0<p<1$. Write the full likelihood, identify a proportional likelihood, and find the MLE of $p$.

Method. Multiply the three binomial probability mass functions. For a binomial count $Y_j$ with size $n_j$, the probability of observing $y_j$ is $\binom{n_j}{y_j}p^{y_j}(1-p)^{n_j-y_j}$.

The full likelihood is therefore

L(p) =\binom{8}{3}\binom{10}{4}\binom{6}{1} p^{3+4+1}(1-p)^{(8-3)+(10-4)+(6-1)}.

The combinatorial factors are $\binom{8}{3}=56$, $\binom{10}{4}=210$, and $\binom{6}{1}=6$. Their product is $56(210)(6)=70{,}560$. There are $3+4+1=8$ observed successes and $8+10+6=24$ trials, so there are $24-8=16$ failures. Thus

L(p)=70{,}560p^8(1-p)^{16} \quad\text{and}\quad L(p)\propto p^8(1-p)^{16}.

The multiplier $70{,}560$ is positive and depends only on the fixed group sizes and observed counts. It does not depend on $p$, so it can be omitted for maximizing. The log of the proportional likelihood is $8\log p+16\log(1-p)$. Its derivative is

\frac{8}{p}-\frac{16}{1-p}=0 \quad\Longrightarrow\quad 8(1-p)=16p \quad\Longrightarrow\quad \widehat p=\frac{8}{24}=\frac{1}{3}.

The second derivative is $-8/p^2-16/(1-p)^2$, which is negative for $0<p<1$; hence this critical point is the maximum. Check: the estimate is the total number of observed successes divided by the total number of trials, and it lies in the allowed interval. The common mistake avoided is treating the binomial coefficients as if they depended on $p$; they do not, so omitting them leaves the MLE unchanged.

Worked Example: Poisson Counts With Fixed Exposures

In a hypothetical study, three fixed exposure periods have lengths $2$, $3$, and $1$ units. The respective observed event counts are $2$, $3$, and $1$. Assume independent Poisson counts with means $\lambda t_j$, where $t_j$ is the known exposure and $\lambda>0$ is the unknown event rate per exposure unit. Find the full and proportional likelihoods and the MLE.

Method. For a Poisson random variable with mean $\lambda t_j$, the probability of count $y_j$ is $e^{-\lambda t_j}(\lambda t_j)^{y_j}/y_j!$. Multiplying the three probabilities gives

L(\lambda) =\prod_{j=1}^{3}\frac{e^{-\lambda t_j}(\lambda t_j)^{y_j}}{y_j!} =e^{-6\lambda}\lambda^6 \frac{2^2\,3^3\,1^1}{2!\,3!\,1!}.

Here the total exposure is $2+3+1=6$, and the total count is $2+3+1=6$. The factor involving only the fixed exposures and observed counts is

\frac{2^2\,3^3\,1}{2!\,3!\,1!} =\frac{4(27)}{2(6)(1)} =\frac{108}{12} =9.

Thus $L(\lambda)=9\lambda^6e^{-6\lambda}$, and a proportional likelihood is $\lambda^6e^{-6\lambda}$. The omitted factor $9$ contains no unknown parameter. Maximizing the log of the proportional likelihood gives

\ell^*(\lambda)=6\log\lambda-6\lambda, \qquad \frac{d\ell^*}{d\lambda}=\frac{6}{\lambda}-6=0 \quad\Longrightarrow\quad \widehat\lambda=1.

The second derivative is $-6/\lambda^2<0$, so the critical point is a maximum. Check: the estimate equals total events divided by total exposure, $6/6=1$ event per exposure unit. The common mistake avoided is dropping the exposure powers without first checking them: they are safe here because each $t_j$ is fixed, but would not be parameter-free if an exposure value itself were unknown.

Worked Example: Normal Mean With Known Variance

A hypothetical calibration study produces independent observations $1$, $2$, and $5$. Assume each observation has a normal distribution with unknown mean $\mu$ and known variance $4$. Find a proportional likelihood and the MLE of $\mu$.

Method. The density of each observation is $(1/\sqrt{8\pi})\exp(-(x_i-\mu)^2/8)$. Independence means the joint density is the product of the three densities:

L(\mu) =\left(\frac{1}{\sqrt{8\pi}}\right)^3 \exp\left(-\frac{(1-\mu)^2+(2-\mu)^2+(5-\mu)^2}{8}\right).

The sum of the observed values is $1+2+5=8$, and the sum of their squares is $1^2+2^2+5^2=30$. Expanding the squared deviations gives

(1-\mu)^2+(2-\mu)^2+(5-\mu)^2 =30-16\mu+3\mu^2.

The density prefactor is fixed, and the factor $\exp(-30/8)$ is also independent of $\mu$. Removing these positive factors leaves

L^*(\mu) =\exp\left(2\mu-\frac{3\mu^2}{8}\right).

The log of this expression is $2\mu-3\mu^2/8$, with derivative $2-3\mu/4$. Setting the derivative to zero gives $\widehat\mu=8/3$. Its second derivative is $-3/4<0$, confirming a maximum. Check: the observed sample mean is $(1+2+5)/3=8/3$, which agrees. The common mistake avoided is assuming that a factor involving the data is always removable: here it is removable because the variance is known and the factor is independent of the only unknown parameter, $\mu$.

When a Term That Looks Constant Is Not

“Constant” must always be understood relative to the unknown parameters. In the preceding normal example, $\exp(-30/8)$ was fixed because the variance was known. If both the mean $\mu$ and variance $\sigma^2$ are unknown, the corresponding full likelihood for the same observations contains

L(\mu,\sigma^2) =(2\pi\sigma^2)^{-3/2} \exp\left(-\frac{30-16\mu+3\mu^2}{2\sigma^2}\right).

The factor $\exp(-30/(2\sigma^2))$ depends on the unknown variance. It therefore cannot be removed when maximizing jointly over $\mu$ and $\sigma^2$. The part $(2\pi)^{-3/2}$ is parameter-free, but $(\sigma^2)^{-3/2}$ is not. This example illustrates why the list of unknown parameters must be checked before simplifying.

For the same observations, the actual joint maximizers are $\widehat\mu=8/3$ and $\widehat{\sigma}^2=26/9$: the sum of squared deviations from $8/3$ is $26/3$, and dividing by the sample size $3$ gives $26/9$. Dropping the data-square term that depends on $\sigma^2$ would not preserve that joint likelihood problem.

Common Mistakes and Exam Checks

Dropping a parameter-dependent factor. A term such as $e^{-6\lambda}$ in a Poisson likelihood is essential because it changes with $\lambda$.
Checking only the parameter currently being solved for. In a likelihood involving $(\mu,\sigma^2)$, a factor independent of $\mu$ may still depend on $\sigma^2$. It cannot be omitted from joint maximization on that basis.
Confusing a data-dependent factor with a parameter-dependent factor. A factorial such as $y!$ is fixed once the count $y$ is observed. A factorial involving an unknown integer parameter would not be fixed in the same way.
Using proportionality for a different purpose without checking. A simplified likelihood is sufficient to locate the MLE, but it does not retain the same numerical likelihood values as the full expression. Keep the full expression when the actual likelihood value is needed.

Key takeaway: You may omit a factor from a likelihood for MLE calculation only when it is positive and independent of every unknown parameter being optimized. Verify that condition from the model and the complete parameter list, not from appearance alone.

Check Your Understanding

For each problem, identify any safe parameter-free factors and use the likelihood to determine the requested result.

A hypothetical group of $12$ independent devices has $5$ observed failures. Assume a common unknown failure probability $p$. Which factor in the binomial likelihood may be omitted, and what is the MLE of $p$? (Answer: omit $\binom{12}{5}=792$; $\widehat p=5/12$.)
Independent Poisson counts are $2$ and $5$, observed over fixed exposures $1$ and $4$, with unknown rate $\lambda$ per exposure unit. Give the full-likelihood factor multiplying $\lambda^7e^{-5\lambda}$, and find the MLE. (Answer: the factor is $1^2 4^5/(2!5!)=64/15$; $\widehat\lambda=7/5=1.4$.)
Independent observations $1,2,3$ come from an exponential distribution with unknown rate $\theta>0$, with density $f(x)=\theta e^{-\theta x}$ for $x>0$. Write the likelihood and find its MLE. (Answer: $L(\theta)=\theta^3e^{-6\theta}$; no additional data-only factor is needed; $\widehat\theta=1/2$.)
Independent normal observations are $-1,2,5$, with unknown mean $\mu$ and known variance $9$. Find a proportional likelihood and the MLE of $\mu$. (Answer: $L^*(\mu)=\exp((6\mu-3\mu^2)/18)$; $\widehat\mu=2$.)
For a multinomial model with three categories, observed counts are $4,3,1$, and the probabilities are $p,q,1-p-q$, where $p>0$, $q>0$, and $p+q<1$. State the factor that may be omitted and the joint MLEs. (Answer: omit $8!/(4!3!1!)=280$; $\widehat p=4/8=1/2$, $\widehat q=3/8$, and the third estimated probability is $1/8$.)

What You'll Learn