Factorizing Likelihoods For Independent Samples

Why Likelihoods Factor

A likelihood for several observations must account for the observations jointly. When the observations are independent, their joint density or probability mass function is a product of their individual contributions. This remains true when observations have different known model settings; identical distributions are not required.

That distinction matters in actuarial work. A set of hypothetical losses might be observed under different known exposure amounts, rating multipliers, or measurement conditions. If the observations are independent, their likelihood contributions multiply, but the contribution for each observation must use its own setting. This tutorial develops that rule and shows how to identify the parts of the product that involve unknown parameters. The methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X_1,\ldots,X_n$ be independent observations with observed values $x_1,\ldots,x_n$. If observation $i$ has density or probability mass function $f_i(x_i;\theta,z_i)$, where $\theta$ is unknown and $z_i$ is a known setting, the likelihood is $$ L(\theta)=\prod_{i=1}^{n} f_i(x_i;\theta,z_i). $$ The observed values and known settings are fixed when the likelihood is viewed as a function of $\theta$. A sample is identically distributed only when the individual distributions are the same; independence alone does not imply this.

For continuous observations, $f_i$ is a density, so the likelihood is not generally the probability of observing the exact data. For discrete observations, it is a probability mass. In either case, the observed data are held fixed and the likelihood varies with the unknown parameter.

A factor that does not involve any unknown parameter may be removed when writing a likelihood up to proportionality. It must be retained if an exact likelihood value is requested. A factor must not be treated as constant if it depends on an unknown parameter, even if it is numerically fixed after a particular parameter value is substituted. Any support condition that affects whether the observation is possible must also be retained.

Why the Product Formula Holds

The product rule is a direct consequence of independence. For independent discrete observations, the joint probability of a particular vector of outcomes is the product of their individual probabilities. For independent continuous observations, the joint density factors in the same way. Substituting the observed values into that joint distribution gives the likelihood.

f_{X_1,\ldots,X_n}(x_1,\ldots,x_n;\theta) = \prod_{i=1}^{n} f_i(x_i;\theta,z_i) \quad\Longrightarrow\quad L(\theta)=\prod_{i=1}^{n} f_i(x_i;\theta,z_i).

This argument also explains what changes when observations are dependent: the joint distribution generally cannot be replaced by the product of marginal contributions. Likewise, if the settings $z_i$ are random rather than fixed, their distribution may contribute to the likelihood. If that distribution depends on $\theta$, it cannot be ignored.

When all observations share the same distribution, the formula simplifies to a product of the same function evaluated at different data values. That is the identically distributed special case, not the general rule. With different known settings, use the appropriate individual model for each observation before multiplying.

Worked Likelihood Factorizations

Worked Example: Three Complete Loss Observations

A hypothetical portfolio has three independent complete loss observations, measured in thousands of dollars: $x_1=0.5$, $x_2=1$, and $x_3=1.5$. Assume each loss has an exponential density with common unknown rate $\lambda>0$: $$ f(x;\lambda)=\lambda e^{-\lambda x},\qquad x\geq 0. $$ Find the likelihood and evaluate it at $\lambda=0.4$.

Method. Use the density contribution for each observed loss, then multiply because the observations are independent.

For $x_1=0.5$, the contribution is $\lambda e^{-0.5\lambda}$. For $x_2=1$, it is $\lambda e^{-\lambda}$. For $x_3=1.5$, it is $\lambda e^{-1.5\lambda}$. Therefore, $$ L(\lambda) =(\lambda e^{-0.5\lambda})(\lambda e^{-\lambda})(\lambda e^{-1.5\lambda}) =\lambda^3 e^{-3\lambda}. $$

At $\lambda=0.4$, $\lambda^3=(0.4)^3=0.064$, and the exponent is $-3(0.4)=-1.2$. Thus $$ L(0.4)=0.064e^{-1.2} =0.064(0.3011942) =0.0192764. $$

Rounded to six decimal places, the likelihood is $0.019276$, in density units corresponding to three observations measured in thousands of dollars. The check is that the power of $\lambda$ is three, one factor from each observation, and the exponent uses the total loss $0.5+1+1.5=3$. The common mistake avoided here is using only one observation's contribution instead of the joint product.

Worked Example: Different Known Exposure Multipliers

Three independent hypothetical loss measurements have known exposure multipliers $e_1=1$, $e_2=2$, and $e_3=4$. Conditional on its multiplier, observation $i$ has an exponential density with rate $\theta e_i$, where $\theta>0$ is unknown. The observed losses, in thousands of dollars, are $x_1=0.5$, $x_2=0.25$, and $x_3=0.125$. Form the likelihood and evaluate it at $\theta=0.2$.

Method. Apply the exponential density using each observation's own known rate. For multiplier $e_i$, the density contribution is $\theta e_i e^{-\theta e_i x_i}$.

The three contributions are $$ \theta e^{-\theta(0.5)},\qquad 2\theta e^{-\theta(2)(0.25)},\qquad 4\theta e^{-\theta(4)(0.125)}. $$

Each product $e_i x_i$ is $0.5$: $1(0.5)=0.5$, $2(0.25)=0.5$, and $4(0.125)=0.5$. The product of the rate multipliers is $1(2)(4)=8$. Therefore, $$ L(\theta) =8\theta^3 e^{-\theta(0.5+0.5+0.5)} =8\theta^3e^{-1.5\theta}. $$

At $\theta=0.2$, $8\theta^3=8(0.008)=0.064$, and $e^{-1.5\theta}=e^{-0.3}=0.7408182$. Hence $$ L(0.2)=0.064(0.7408182)=0.0474124. $$

Rounded to six decimal places, the likelihood is $0.047412$. The calculation checks because all three rates use the same unknown baseline parameter but different known multipliers. The common mistake avoided is treating the observations as identically distributed and omitting the multipliers $1,2,4$; doing so would lose the factor $8$ and would also use the wrong exponent.

Worked Example: Successes and Failures Under Different Known Settings

A hypothetical insurer independently records whether three contracts produce a claim indicator $X_i$, with $X_i=1$ for a claim and $X_i=0$ otherwise. For contract $i$, the known rating multiplier is $r_i$, and the model is $$ P(X_i=1\mid \theta,r_i)=\frac{\theta r_i}{1+\theta r_i}, \qquad P(X_i=0\mid \theta,r_i)=\frac{1}{1+\theta r_i}, \qquad \theta>0. $$ The multipliers are $1,2,\tfrac12$, and the observed outcomes are $1,0,1$. Find the likelihood and evaluate it at $\theta=1$.

Method. Use a success probability for each observed 1 and a failure probability for the observed 0, retaining the distinct known multiplier in each factor.

For the first contract, $r_1=1$ and $X_1=1$, so the contribution is $\theta/(1+\theta)$. For the second, $r_2=2$ and $X_2=0$, so the contribution is $1/(1+2\theta)$. For the third, $r_3=\tfrac12$ and $X_3=1$, so the contribution is $$ \frac{\theta/2}{1+\theta/2}=\frac{\theta}{2+\theta}. $$

Multiplying the three independent contributions gives $$ L(\theta) =\frac{\theta}{1+\theta} \cdot\frac{1}{1+2\theta} \cdot\frac{\theta}{2+\theta} =\frac{\theta^2}{(1+\theta)(1+2\theta)(2+\theta)}. $$

At $\theta=1$, the numerator is $1$, and the denominator is $(2)(3)(3)=18$. Thus $L(1)=1/18=0.0556$ to four decimal places. This is a probability mass, not a density. The check is that the middle observation is a failure, so its factor is $1/(1+2\theta)$, not the success probability. The common mistake avoided is using success contributions for every record merely because the model is specified by a success probability.

Parameter-Dependent and Parameter-Free Factors

A useful final step in a factorization is to mark which parts vary with the unknown parameter. For the exposure example, the exact likelihood is $8\theta^3e^{-1.5\theta}$. Since $8$ is known and does not involve $\theta$, it can be omitted if the requested result is only a likelihood proportional to the full likelihood: $$ L(\theta)\propto \theta^3e^{-1.5\theta}. $$ The omitted factor must be restored to report the exact likelihood value. In the indicator example, all three denominator factors involve $\theta$; none can be discarded as a parameter-free constant.

Known settings can create constants and parameter-dependent terms in the same contribution. For example, a density may include a known multiplier outside an exponential term while the unknown parameter appears both in the rate and in the exponent. Inspect the entire factor rather than deciding that a contribution is constant merely because its setting is known.

A support restriction requires particular care. If a model assigns zero density to some observed value for certain parameter values, the likelihood is zero there. It is not valid to drop that restriction when simplifying the likelihood. Likelihoods that differ by a positive factor independent of all unknown parameters have the same relative comparisons across parameter values, but replacing a likelihood with a proportional expression does not change what values of the parameter are permitted by the data and model.

State the model and independence assumption.
Write the distribution for each observation, including any known setting that changes its contribution.

Substitute the observed values.
Each factor must correspond to the actual outcome and setting for that observation.

Multiply and simplify.
Collect repeated parameter factors and combine exponents or other terms only after writing the individual contributions.

Check the result.
Confirm that every observation contributed once, the settings are correctly matched, and no parameter-dependent factor or support condition was discarded.

Common Exam Traps

Assuming independence means identical distributions. Independence justifies multiplication. It does not justify replacing different known settings by a common one.
Forgetting an outcome type. For a binary observation, use the probability of the outcome actually observed; a failure contributes $1-p$, not $p$.
Dropping a factor that contains the parameter. A known multiplier may be constant, but a term such as $e^{-\theta c}$ still depends on $\theta$.
Reporting a proportional likelihood as an exact value. Parameter-free factors may be removed for proportional comparisons, but not when an exact likelihood value is requested.
Multiplying marginal contributions for dependent observations. Independence is an assumption needed for this factorization; without it, use the joint model.

Key takeaway: For independent observations, multiply the correct individual contributions, using each observation's own known settings. Simplify only after forming the product, and distinguish an exact likelihood from one stated up to a parameter-free proportionality factor.

Check Your Understanding

For each problem, form the likelihood from the stated independent observations and pay attention to which factors involve unknown parameters.

Two independent losses have exponential density $f(x;\lambda)=\lambda e^{-\lambda x}$, for $\lambda>0$. The observed losses are $1$ and $2$. Give the likelihood and its value at $\lambda=0.5$, rounded to six decimal places. (Answer: $L(\lambda)=\lambda^2e^{-3\lambda}$; $L(0.5)=0.055783$.)
Two independent hypothetical losses have exponential rates $\theta c_i$, with known multipliers $c_1=1$, $c_2=3$, and observations $x_1=2$, $x_2=1$. Give the exact likelihood. (Answer: $L(\theta)=3\theta^2e^{-5\theta}$.)
Two independent indicators have success probabilities $\theta/(\theta+c_i)$, with known $c_1=1$, $c_2=3$. The observed outcomes are failure then success. Give the likelihood. (Answer: $L(\theta)=\theta/[(1+\theta)(3+\theta)]$.)
Independent observations satisfy $X_i\sim N(\theta+a_i,\sigma_i^2)$, where $a_1=1$, $a_2=-1$, $\sigma_1^2=4$, and $\sigma_2^2=1$; the observations are $x_1=3$, $x_2=1$. Give the likelihood as a function of $\theta$. (Answer: $L(\theta)=\frac{1}{4\pi}\exp[-\frac{5}{8}(2-\theta)^2]$.)
Three independent losses have exponential rates $\theta c_i$, with known multipliers $(c_1,c_2,c_3)=(2,1,3)$ and observed losses $(x_1,x_2,x_3)=(0.25,0.5,0.1)$. Give both the exact likelihood and a proportional likelihood. (Answer: $L(\theta)=6\theta^3e^{-1.3\theta}$; $L(\theta)\propto\theta^3e^{-1.3\theta}$.)

What You'll Learn