Likelihoods From Complete Loss Observations

Why Likelihoods Start With the Observations

A loss model specifies how observations could arise under a particular parameter value. When exact losses are observed, the likelihood records how compatible each possible parameter value is with the data. It is a basic building block of parametric estimation: before estimating a parameter, one must express the likelihood correctly.

The essential operation is direct. For independent observations, use the model's probability mass at each observed value if the model is discrete, or its density at each observed value if the model is continuous. Multiply those contributions to obtain the joint probability or density of the observed data. Then regard that expression as a function of the unknown parameter.

Definition: Suppose $X_1,\ldots,X_n$ are independent observations from a model indexed by parameter $\theta$, and their exact observed values are $x_1,\ldots,x_n$. The likelihood is the joint probability mass (for discrete data) or joint density (for continuous data), evaluated at the observations and written as a function of $\theta$: $$ L(\theta;x_1,\ldots,x_n)=\prod_{j=1}^{n} f(x_j;\theta), $$ where $f$ denotes a probability mass function or probability density function as appropriate. For discrete observations, $f(x_j;\theta)$ is a probability; for continuous observations, it is a density. The formula requires independence and a correctly specified model.

The expression $L(\theta;x_1,\ldots,x_n)$ depends on the data and the parameter in different ways. In likelihood calculations, the observed values are fixed and $\theta$ varies. This is not a claim that the parameter itself is random. A likelihood value is not, by itself, the probability that $\theta$ is true.

Masses, Densities, and Support

For a discrete loss model, the contribution from an observed value is its probability mass. For example, if $P(X=x;\theta)$ is the probability that a loss equals $x$, then the contribution for an exact observation $x$ is $P(X=x;\theta)$. It is a probability between zero and one.

For a continuous loss model, an exact value has probability zero: $P(X=x;\theta)=0$. The contribution is instead the density $f_X(x;\theta)$. A density can be greater than one, and it is not the probability of observing that exact loss. The probability of an interval is found by integrating the density across the interval.

Key distinction: The likelihood uses the model's probability mass or density at the observed value. It does not use the probability of a continuous observation equaling that exact value, which is zero. In either case, the likelihood is a function of the parameter after the data have been fixed.

The model's support—the set of values that can occur—must be respected. If an observed loss lies outside the support for a parameter value, that parameter value assigns the data probability or density zero, so its likelihood is zero. Conversely, if the data are within the support, use the stated mass or density at each observation.

When the observations are independent and have the same model, each contribution uses the same parameterized mass or density. The product is the joint mass or density of the observed sequence. The product rule follows directly from independence: the joint probability of independent discrete events is the product of their probabilities, and the joint density of independent continuous variables is the product of their marginal densities.

Identify the data type and model.
Determine whether each exact observation is discrete or continuous, and state the model and parameter restrictions.

Check support.
For each candidate parameter value, verify that every observed loss is permitted by the model.

Write each contribution.
Use a probability mass for a discrete observation or a density for a continuous observation.

Multiply and identify the function.
Multiply the contributions, then treat the observed values as fixed and the parameter as variable.

Worked Likelihoods

The examples use hypothetical claim data. Each states the model assumptions and shows how to move from individual contributions to a likelihood function. The calculations construct likelihoods; they do not require finding a maximum likelihood estimate.

Worked Example: A Discrete Loss Amount

A hypothetical repair plan records each claim as one of three exact amounts: $100, $200, or $500. Claims are independent and follow this model, where $\theta$ is a parameter satisfying $0<\theta<1/3$:

P(X=100;\theta)=\theta,\qquad P(X=200;\theta)=2\theta,\qquad P(X=500;\theta)=1-3\theta.

The observed ordered sequence is $100, $200, $500, $500. Construct its likelihood and compare the values at $\theta=0.10$ and $\theta=0.20$.

Step 1: Check the model. For $0<\theta<1/3$, all three probabilities are positive, and their sum is $\theta+2\theta+(1-3\theta)=1$. The data consist of exact discrete amounts, so each contribution is a probability mass.

Step 2: Write each contribution. In order, the four observations contribute $\theta$, $2\theta$, $1-3\theta$, and $1-3\theta$. Independence makes the joint probability their product:

L(\theta)=\theta(2\theta)(1-3\theta)(1-3\theta) =2\theta^2(1-3\theta)^2.

Step 3: Evaluate at $\theta=0.10$. The four masses are $0.10$, $0.20$, $0.70$, and $0.70$, so $L(0.10)=0.10(0.20)(0.70)(0.70)=0.0098$.

Step 4: Evaluate at $\theta=0.20$. The masses are $0.20$, $0.40$, $0.40$, and $0.40$, so $L(0.20)=0.20(0.40)(0.40)(0.40)=0.0128$.

Thus the likelihood at $0.20$ is larger for this observed sequence than the likelihood at $0.10$. This comparison does not say that $0.20$ is the probability of the parameter. Method and trap avoided: multiply the masses for the ordered observations; do not count the observed sequence as unordered and add a combinatorial coefficient that is not part of its joint probability.

Worked Example: A Continuous Model With Parameter-Dependent Support

Suppose three independent hypothetical claim severities, measured in thousands of dollars, follow a uniform distribution on $(0,\theta)$, with $\theta>0$. Their exact observed values are $2$, $3$, and $4$. Construct the likelihood and evaluate it at $\theta=4$ and $\theta=5$.

Step 1: State the density and support. For this model,

f_X(x;\theta)= \begin{cases} 1/\theta, & 0<x<\theta,\\ 0, & \text{otherwise}. \end{cases}

The largest observation is $4$. If $\theta\leq 4$, at least one observation is outside the model's support, so the likelihood is zero. If $\theta>4$, all three density contributions equal $1/\theta$. Therefore,

L(\theta)= \begin{cases} 0, & \theta\leq 4,\\ \theta^{-3}, & \theta>4. \end{cases}

Step 2: Evaluate at $\theta=4$. Under the stated open support $(0,\theta)$, the observation $x=4$ is not in the support when $\theta=4$. Thus $L(4)=0$.

Step 3: Evaluate at $\theta=5$. Each observation is in the support, and each density is $1/5$. Hence $L(5)=(1/5)(1/5)(1/5)=1/125=0.008$, in units of inverse thousands-of-dollars cubed.

The likelihood is a product of densities, not the probability of the exact triple $(2,3,4)$, which is zero for continuous observations. Method and trap avoided: check the support before applying the density formula; using $1/\theta$ at $\theta=4$ would ignore the specified open endpoint and incorrectly give a positive likelihood.

Worked Example: Exponential Densities at Candidate Parameter Values

A hypothetical portfolio has three independent exact claim severities of $1$, $2$, and $3$ thousand dollars. Assume each severity has an exponential distribution with mean $\theta>0$, so its density is $f_X(x;\theta)=\theta^{-1}e^{-x/\theta}$ for $x>0$. Construct the likelihood and evaluate it at $\theta=2$ and $\theta=3$, with $\theta$ also measured in thousands of dollars.

Step 1: Confirm support and contributions. All three observed values are positive, as required by the exponential model. Their density contributions are $\theta^{-1}e^{-1/\theta}$, $\theta^{-1}e^{-2/\theta}$, and $\theta^{-1}e^{-3/\theta}$.

Step 2: Multiply. Multiplying the three factors gives three factors of $\theta^{-1}$, and the exponential terms combine because their exponents add:

L(\theta)=\theta^{-3}e^{-(1+2+3)/\theta} =\theta^{-3}e^{-6/\theta},\qquad \theta>0.

Step 3: Evaluate at $\theta=2$. Here $\theta^{-3}=1/8=0.125$ and $e^{-6/2}=e^{-3}\approx0.049787$. Thus $L(2)=0.125(0.049787)\approx0.006223$, in units of inverse thousands-of-dollars cubed.

Step 4: Evaluate at $\theta=3$. Here $\theta^{-3}=1/27\approx0.037037$ and $e^{-6/3}=e^{-2}\approx0.135335$. Thus $L(3)=(1/27)(0.135335)\approx0.005012$, in the same units.

The likelihood is larger at $\theta=2$ than at $\theta=3$ for these data. The calculations are rounded to six decimal places. Method and trap avoided: use the mean parameterization given in the problem and sum the observed losses in the exponent; confusing the mean with a rate would produce a different and incorrect function.

Why the Product Rule Holds

For discrete independent observations, let $A_j$ be the event that observation $j$ equals its recorded value $x_j$. Independence means the joint probability of these events is the product of their individual probabilities. Therefore,

P(X_1=x_1,\ldots,X_n=x_n;\theta) =\prod_{j=1}^{n}P(X_j=x_j;\theta).

For continuous independent observations, independence gives a joint density equal to the product of marginal densities. A small-interval argument gives the same intuition: for small widths $h_1,\ldots,h_n$, the probability that each $X_j$ falls in $[x_j,x_j+h_j]$ is approximately $\prod_{j=1}^{n}f(x_j;\theta)h_j$. Dividing by the product of interval widths and taking the widths toward zero yields the joint density $\prod_{j=1}^{n}f(x_j;\theta)$. In both cases, evaluating the joint mass or density at the observed data and then varying $\theta$ produces the likelihood.

Independence is essential to this product formula. If observations are dependent, their joint mass or joint density must be used instead; multiplying marginal contributions alone generally does not give the likelihood. Likewise, incomplete observations require contributions that reflect what was actually observed, rather than treating an unobserved exact loss as if it were known.

Likelihood Is Not Probability About the Parameter

A probability model describes possible data for a fixed parameter. A likelihood reverses the viewpoint: after data are observed, it compares how well different fixed parameter values account for those data. For example, if $L(0.20)>L(0.10)$, the observed data have a greater joint probability or density under $0.20$ than under $0.10$. This statement is a comparison of likelihoods, not a probability assignment to the parameter values.

A common incorrect calculation is to take likelihood values and call them probabilities that the parameter is true. In the discrete example, one might incorrectly say, “The probability that $\theta=0.20$ is $0.0128$.” That is wrong: $0.0128$ is the probability of the particular observed sequence under the model with $\theta=0.20$. It is a likelihood value when viewed as a function of $\theta$. A probability distribution for an unknown parameter requires additional modeling, such as a prior distribution; it does not follow from the likelihood alone.

Another useful distinction is between the likelihood function and a single likelihood value. The function $L(\theta;x_1,\ldots,x_n)$ displays the value for every permitted $\theta$. Substituting a particular value gives one point on that function. Likelihoods need not sum or integrate to one across parameter values, and they are not required to lie between zero and one when densities are involved.

Common Checks Before Moving On

Exact data: use the mass or density for the exact observed value. Do not substitute an interval probability unless the observation was reported as an interval.
Correct model: state the distribution and its parameterization. A mean and a rate, for example, are not interchangeable parameter labels.
Support: a parameter value that cannot generate the observed data has likelihood zero.
Independence: multiply individual contributions only when the observations are independent under the model.
Interpretation: a likelihood compares parameter values for fixed data; it is not a probability distribution over parameters.

Key takeaway: For exact independent observations, construct the likelihood by multiplying each observation's model mass or density, enforce the model's support, and then regard the resulting joint expression as a function of the parameter. Keep the data fixed and do not interpret likelihood as probability about the parameter.

Check Your Understanding

For each problem, identify the likelihood and pay attention to whether the data are discrete or continuous.

A hypothetical claim indicator takes values $0$ and $1$, with $P(X=1;p)=p$ and $P(X=0;p)=1-p$. Three independent exact observations are $1,0,1$. Write the likelihood. (Answer: $L(p)=p^2(1-p)$, for $0\leq p\leq1$.)
Two independent hypothetical losses have a uniform distribution on $(0,\theta)$, with $\theta>0$, and are observed as $1.5$ and $2.5$ thousand dollars. Give the likelihood and its value at $\theta=2$ and $\theta=3$. (Answer: $L(\theta)=0$ for $\theta\leq2.5$, and $L(\theta)=\theta^{-2}$ for $\theta>2.5$; $L(2)=0$, $L(3)=1/9$.)
Two independent exact severities are $0.5$ and $1.5$ thousand dollars from an exponential distribution with mean $\theta>0$, also in thousands. Write the likelihood and evaluate it at $\theta=1$, to four decimals. (Answer: $L(\theta)=\theta^{-2}e^{-2/\theta}$; $L(1)=e^{-2}=0.1353$, in inverse thousands-of-dollars squared.)
A discrete hypothetical loss takes values $10$ and $20$, with probabilities $P(X=10;\theta)=\theta$ and $P(X=20;\theta)=1-\theta$, where $0<\theta<1$. The independent observed sequence is $20,10,10$. Write its likelihood and evaluate it at $\theta=0.4$. (Answer: $L(\theta)=(1-\theta)\theta^2$; $L(0.4)=0.6(0.4)^2=0.096$.)

What You'll Learn