Why Simulate Choice Probabilities?
In a discrete-choice model, the probability of an observed outcome may involve averaging over a latent variable that is not observed. That average can be difficult to evaluate analytically, even when the probability conditional on the latent variable is simple. Simulation replaces the difficult average with a finite average over draws. Simulated maximum likelihood (SML) then uses those approximated probabilities in a likelihood and chooses parameter values that make the observed outcomes relatively likely.
The order of operations matters: first average the conditional probabilities over latent heterogeneity, then take the logarithm and add contributions across observations. Averaging conditional log probabilities instead generally gives a different objective and does not approximate the likelihood being specified.
From Latent Heterogeneity to a Simulated Probability
This finite average approximates the latent-variable integral. If the latent distribution depends on \(\theta\), draws can be generated by transforming draws from a parameter-free base distribution; then the simulated probability is evaluated using the transformed draws at each parameter value. The examples below use a fixed, parameter-free latent distribution so the construction is especially transparent.
For a binary outcome, it is sufficient to approximate the success probability \(p_i(\theta)\). The failure probability is \(1-p_i(\theta)\), and using the same draws for the two outcomes preserves that relationship. A probability used inside a logarithm must be positive; in this example each conditional probability lies strictly between zero and one, so every finite average does as well.
Worked Example: Averaging Conditional Choice Probabilities
Consider a hypothetical binary-choice model. Each person has an unobserved type \(v\), equal to \(+1\) or \(-1\) with probability one-half each. Conditional on type, the probability of choosing option 1 is $$ q(\theta,v)=\frac{1}{1+\exp[-(\theta+v)]}. $$ For a person with the same observed covariates as the others, the population choice probability is the average of the two type-specific probabilities. At \(\theta=0.8\), the two values are $$ q(0.8,+1)=\frac{1}{1+\exp(-1.8)}\approx 0.8581489,\qquad q(0.8,-1)=\frac{1}{1+\exp(0.2)}\approx 0.4501660. $$
Use one draw of each type, \(v_1=+1\) and \(v_2=-1\). The simulated probability is $$ \widehat p_{2}(0.8)=\frac{0.8581489+0.4501660}{2}=0.6541575. $$ Because these two draws represent the two possible types in exactly their population proportions, this particular finite average equals the exact integrated probability, up to rounding. With random draws that do not represent the two types equally, the finite average would instead approximate the exact probability with simulation error.
Constructing the Simulated Log Likelihood
Suppose the observations are independent conditional on the model parameters and observed covariates. The likelihood is the product of their outcome probabilities, and the log likelihood is the sum of their log probabilities. SML substitutes the simulated probabilities for the probabilities that require integration.
In a binary model, an observation with \(y_i=1\) contributes \(\log\widehat p_{i,R}(\theta)\); an observation with \(y_i=0\) contributes \(\log[1-\widehat p_{i,R}(\theta)]\). This distinction is essential: the contribution is the log of the simulated probability of the outcome actually observed.
Worked Example: Calculating the Simulated Log Likelihood
Suppose four hypothetical people have identical observed covariates, with outcomes \(1,1,1,0\). At \(\theta=0.8\), use the two balanced draws from the preceding example, so that \(\widehat p_2(0.8)=0.6541575\). The simulated log likelihood is $$ \widehat\ell_2(0.8) =3\log(0.6541575)+\log(1-0.6541575) \approx -2.334993. $$
The first term is the contribution from the three people who chose option 1. The second is the contribution from the person who chose option 0, whose simulated probability is \(1-0.6541575=0.3458425\). The negative value is expected because logarithms of probabilities strictly between zero and one are negative. It is the relative size of the log likelihood across parameter values, not its sign alone, that matters for maximization.
Draw Reuse and the Simulated Objective
A practical SML calculation typically holds the latent-variable draws fixed while evaluating the objective at different parameter values. This is called draw reuse. If each candidate parameter value instead receives a fresh random sample, the objective can jump around partly because the draws changed, making it harder to distinguish parameter effects from simulation variation. Fixed draws do not eliminate approximation error, but they make comparisons across parameter values reproducible.
Here is a second draw set for the hypothetical two-type model: \(v_1=v_2=v_3=+1\) and \(v_4=-1\). At \(\theta=0.8\), reusing these four draws gives $$ \widehat p_4(0.8) = \frac{3(0.8581489)+0.4501660}{4} \approx 0.7561532. $$ For the same outcomes \(1,1,1,0\), the simulated log likelihood is $$ \widehat\ell_4(0.8) = 3\log(0.7561532)+\log(1-0.7561532) \approx -2.2497. $$ This differs from the balanced two-draw calculation because the four-draw set places three-quarters, rather than one-half, of its weight on type \(+1\). It is a different approximation to the same population probability, not a different observed outcome data set.
The same four draws can be reused at any trial value of \(\theta\): evaluate \(q(\theta,v_r)\) for each fixed \(v_r\), average those values, and then calculate the log likelihood. If the draws are resampled at every trial value, the resulting changes in the objective mix parameter changes with Monte Carlo variation. Draw reuse makes the comparison cleaner, although a finite draw set can still approximate the integral imperfectly.
Maximizing a Small Simulated Likelihood
The following calculation shows how maximizing the simulated objective works in this example. It also gives a short justification for the target probability in the optimization. For four outcomes \(1,1,1,0\) sharing the same success probability \(p\), the log likelihood as a function of that probability is $$ \ell(p)=3\log(p)+\log(1-p),\qquad 0<p<1. $$ Differentiating, $$ \frac{d\ell(p)}{dp}=\frac{3}{p}-\frac{1}{1-p} =\frac{3-4p}{p(1-p)}. $$ The denominator is positive on \(0<p<1\), so the derivative is positive when \(p<0.75\), zero at \(p=0.75\), and negative when \(p>0.75\). Also, $$ \frac{d^2\ell(p)}{dp^2} =-\frac{3}{p^2}-\frac{1}{(1-p)^2}<0. $$ Thus the log likelihood has a unique maximum at \(p=0.75\). SML chooses \(\theta\) to make its simulated probability as close to this likelihood-maximizing probability as the model and draw set permit.
Worked Example: Maximizing with Reused Draws
For a transparent comparison, use the balanced draws \(+1,-1\), reused at every value of \(\theta\). The simulated probability is $$ \widehat p_2(\theta) = \frac{1}{2}\left[ \frac{1}{1+\exp[-(\theta+1)]} + \frac{1}{1+\exp[-(\theta-1)]} \right]. $$ It increases with \(\theta\), since each conditional logistic probability increases with \(\theta\). The likelihood is therefore maximized when \(\widehat p_2(\theta)=0.75\), if that probability is attainable.
For example, at \(\theta=0\), the two conditional probabilities are \(0.7310586\) and \(0.2689414\), giving \(\widehat p_2(0)=0.5000000\). The log likelihood is \(4\log(0.5)\approx-2.772589\). At \(\theta=0.8\), the probability is \(0.6541575\), and the log likelihood is \(-2.334993\), as calculated above. Both probabilities are below \(0.75\), so increasing \(\theta\) in this range increases the likelihood.
Solving \(\widehat p_2(\theta)=0.75\) gives an approximate maximizer of \(\widehat\theta\approx1.35\). For a numerical check, at \(\theta=1.35\) the conditional probabilities are approximately \(0.9129\) and \(0.5866\), whose average is about \(0.7498\). The corresponding log likelihood is about $$ 3\log(0.7498)+\log(1-0.7498)\approx -2.249. $$ The small difference from the maximum possible value reflects rounding \(\theta\) to two decimal places. This is a simulated-likelihood estimate for the stated fixed draw design; a different finite draw set can produce a different approximation and hence a different estimate.
Common Pitfall and Practical Meaning
A common mistake is to average the conditional log probabilities, using $$ \frac{1}{R}\sum_{r=1}^{R}\log P(Y_i=y_i\mid x_i,v_{ir};\theta), $$ as though it were the simulated log likelihood contribution. The required contribution is instead $$ \log\left[\frac{1}{R}\sum_{r=1}^{R}P(Y_i=y_i\mid x_i,v_{ir};\theta)\right]. $$ These expressions are not generally equal. The first averages log probabilities across latent draws; the second approximates the integrated outcome probability and then takes its logarithm. The latter follows from the likelihood for the observed outcome after latent heterogeneity has been integrated out.
SML matters because it makes likelihood-based estimation feasible when latent-variable probabilities are inconvenient to integrate exactly. But the estimate depends on the approximation used to construct those probabilities. The sample size and the number and design of the simulation draws play different roles: more observed people do not, by themselves, make a particular finite-draw probability approximation more accurate. In applications, the draw design and simulation accuracy must therefore be considered alongside the fitted likelihood.
Check Your Understanding
Answer each question by referring to the model, probability construction, or likelihood calculation described above.
- Why does a binary-outcome model need only a simulated success probability to calculate both outcome probabilities?
- For outcomes \(1,1,0\) with a common simulated success probability \(p\), write the simulated log likelihood.
- Explain why averaging conditional log probabilities is generally not a substitute for taking the logarithm of the averaged conditional probabilities.
- What is the practical advantage of reusing the same latent-variable draws at each candidate parameter value?
- In the four-person example, why did the two-draw and four-draw sets give different simulated probabilities at \(\theta=0.8\)?