Hotz-Miller Inversion of Choice Probabilities

Why Invert Choice Probabilities?

In a dynamic discrete-choice model, a decision maker chooses among alternatives after observing choice-specific shocks. Researchers may be able to describe choice probabilities conditional on the state, while the underlying choice-specific values are not directly observed. The Hotz-Miller inversion connects the two: under Type I extreme-value shocks, relative choice probabilities give differences in choice-specific values directly.

The word relative matters. The probabilities identify how much more valuable one alternative is than a reference alternative, not the absolute level of every value. And the recovered object is a choice-specific value, which includes the continuation value of future states as well as any current payoff.

Model and Assumptions

Let $x$ denote the observed state before a choice, and let $j\in\{0,1,\ldots,J\}$ index the available alternatives. The deterministic choice-specific value $v_j(x)$ is the value of choosing $j$, conditional on state $x$, before the current choice shock is realized. The decision maker chooses the alternative with the highest shock-adjusted value.

Definition: In the normalized model, the choice rule is $$ j^*(x)=\arg\max_{j\in\{0,\ldots,J\}}\{v_j(x)+\varepsilon_j\}, $$ where the shocks $\varepsilon_j$ are independent Type I extreme-value random variables with cumulative distribution function $F(\varepsilon)=\exp[-\exp(-\varepsilon)]$. The conditional choice probability is $P_j(x)=\Pr(j^*(x)=j\mid x)$. The shock scale is normalized to one.

The dynamic content is contained in $v_j(x)$. For example, if $u_j(x)$ is the current payoff and the state next period follows the transition distribution conditional on $x$ and $j$, then a typical choice-specific value is

v_j(x)=u_j(x)+\beta\,\mathbb{E}\!\left[\bar V(x')\mid x,j\right],

where $\beta$ is the discount factor, $x'$ is the next state, and $\bar V(x')$ is the value integrated over next period’s choice shocks. The transition distribution and current payoff may differ across alternatives. The inversion below recovers differences in $v_j(x)$, not $u_j(x)$ alone.

The key assumptions for the standard formula are that the alternatives’ shocks are independent and identically distributed Type I extreme value, and that the deterministic values are well-defined conditional on the observed state. The shock specification must apply to the choice occasion being analyzed; dependence across alternatives, for example, generally changes the choice-probability formula. The Type I extreme-value integration result gives

P_j(x)=\frac{\exp(v_j(x))}{\sum_{k=0}^{J}\exp(v_k(x))}.

This is the multinomial logit choice-probability formula. It applies to population conditional probabilities. In empirical work, probabilities are estimated and may be imprecise, but that estimation issue does not change the population identity.

The Hotz-Miller Inversion

Choose alternative $0$ as a reference. Divide the probability of choosing alternative $j$ by the probability of choosing alternative $0$. The denominator of the logit formula is the same for both alternatives and cancels:

\frac{P_j(x)}{P_0(x)} = \frac{\exp(v_j(x))}{\exp(v_0(x))} = \exp\!\bigl(v_j(x)-v_0(x)\bigr).

Taking logarithms yields the inversion.

Theorem: Under the normalized Type I extreme-value choice model, if $P_j(x)>0$ and $P_0(x)>0$, then for every alternative $j$, $$ v_j(x)-v_0(x)=\log P_j(x)-\log P_0(x) =\log\!\left(\frac{P_j(x)}{P_0(x)}\right). $$ Thus, conditional choice probabilities identify every choice-specific value difference relative to alternative $0$.

This is a complete inversion for relative values: once the probabilities are known, the right-hand side is determined. It also works between any two alternatives $j$ and $k$, since $v_j(x)-v_k(x)=\log(P_j(x)/P_k(x))$. The reference alternative is a normalization for reporting differences, not a claim that it has zero value.

A short justification also shows why absolute values cannot be recovered. Adding the same number $c(x)$ to every $v_j(x)$ leaves each probability unchanged:

\frac{\exp(v_j(x)+c(x))} {\sum_{k=0}^{J}\exp(v_k(x)+c(x))} = \frac{\exp(c(x))\exp(v_j(x))} {\exp(c(x))\sum_{k=0}^{J}\exp(v_k(x))} = P_j(x).

Consequently, probabilities alone cannot distinguish the original values from values shifted by a common state-specific amount. The inversion recovers differences, which are invariant to that shift.

Worked Applications

Worked Example: Inverting a Binary Choice Probability

At a particular state $x$, suppose a hypothetical population chooses alternative $1$ with probability $0.65$ and alternative $0$ with probability $0.35$. Assume the normalized Type I extreme-value model applies. Find the relative choice-specific value.

Both probabilities are positive and sum to one, so the inversion is applicable. Using alternative $0$ as the reference,

v_1(x)-v_0(x) = \log\!\left(\frac{P_1(x)}{P_0(x)}\right) = \log\!\left(\frac{0.65}{0.35}\right) = \log(1.8571) \approx 0.6190.

Thus alternative $1$'s deterministic choice-specific value is about $0.6190$ normalized utility units above alternative $0$'s at this state. This is a difference in values, not a probability and not a difference in current payoffs alone. As a check, the implied probability for alternative $1$ is $\exp(0.6190)/(1+\exp(0.6190))$, approximately $0.65$.

Worked Example: Three Alternatives and a Reference Value

Suppose hypothetical conditional choice probabilities at state $x$ are $P_0(x)=0.20$, $P_1(x)=0.50$, and $P_2(x)=0.30$. Calculate both non-reference values relative to $v_0(x)$, then verify that these differences reproduce the probabilities.

The probabilities are positive and sum to $1.00$, as required. The inversion gives

v_1(x)-v_0(x)=\log\!\left(\frac{0.50}{0.20}\right) =\log(2.5)\approx 0.9163,

v_2(x)-v_0(x)=\log\!\left(\frac{0.30}{0.20}\right) =\log(1.5)\approx 0.4055.

For verification, normalize $v_0(x)$ to zero when computing probabilities. The relative exponentiated values are $1$, $\exp(0.9163)\approx2.5$, and $\exp(0.4055)\approx1.5$. Their sum is $5.0$. Dividing each by the sum gives $1/5=0.20$, $2.5/5=0.50$, and $1.5/5=0.30$, exactly the specified shares up to rounding.

Setting $v_0(x)=0$ here is only a convenient normalization for reconstructing the shares. The probabilities do not establish that the actual value of alternative $0$ is zero.

Worked Example: Separating Current Payoff from Continuation Value

Consider two alternatives, $A$ and $B$, at a hypothetical state. Suppose their conditional choice probabilities are $P_A=0.60$ and $P_B=0.40$. Let the discount factor be $\beta=0.90$, and suppose the expected integrated next-period values are $4$ after choosing $A$ and $3$ after choosing $B$. Find the difference in current payoffs $u_A-u_B$.

First invert the probabilities to obtain the difference in total choice-specific values:

v_A-v_B = \log\!\left(\frac{0.60}{0.40}\right) = \log(1.5) \approx 0.4055.

Under the stated dynamic value decomposition, $v_A=u_A+0.90(4)$ and $v_B=u_B+0.90(3)$. Therefore,

$$ u_A-u_B = (v_A-v_B)-0.90(4-3) = 0.4055-0.9000 = -0.4945. $$

The implied current payoff for $A$ is about $0.4945$ utility units lower than for $B$, even though $A$ is chosen more often at this state. That is consistent with the data as specified: $A$'s one-unit advantage in expected next-period value, discounted to $0.9000$, more than offsets its lower current payoff. The calculation assumes the stated continuation-value expectations and common discount factor are appropriate for these alternatives.

Scale, Positivity, and Interpretation

The scale normalization is substantive. If shocks instead have a known Type I extreme-value scale $\sigma>0$, so that the choice-specific random component is $\sigma\varepsilon_j$ for a standard Type I extreme-value $\varepsilon_j$, then

P_j(x)=\frac{\exp(v_j(x)/\sigma)} {\sum_{k=0}^{J}\exp(v_k(x)/\sigma)} \quad\text{and}\quad v_j(x)-v_0(x) = \sigma\log\!\left(\frac{P_j(x)}{P_0(x)}\right).

A probability ratio therefore identifies a value difference relative to the shock scale. If the scale is not fixed or known, probabilities alone do not establish the numerical scale of utility. In the normalized model used above, $\sigma=1$, so the log probability ratio is expressed in normalized utility units.

For any finite deterministic values under this model, every alternative has strictly positive probability. A zero probability in an empirical table may reflect a finite sample with no observed choices of that alternative, rather than a population probability of exactly zero. The logarithm of zero is undefined, so the inversion cannot be applied directly to such a sample share. One must distinguish an estimated zero from the population condition required by the identity; replacing or smoothing estimated probabilities is an empirical modeling decision, not part of the mathematical inversion.

The recovered differences can be useful because dynamic models often impose structure on how current payoffs, transitions, and continuation values combine. The identity itself does not identify those components separately, nor does it establish that a specified structural model is correct. It supplies relative choice-specific values implied by conditional probabilities under the maintained shock specification.

Common Pitfall

Common mistake: “Alternative $1$ is chosen with probability $0.65$, so its value is $0.65$.” This confuses a choice probability with a choice-specific value. The correct normalized-model calculation compares probabilities: $v_1-v_0=\log(P_1/P_0)$. A probability of $0.65$, by itself, is not a value level and does not reveal a value difference without the reference alternative’s probability.

A related error is to interpret the inversion as recovering the current payoff difference. In a dynamic model, choice-specific values include continuation values. Recovering current payoffs requires additional information about the discount factor, transitions, and future values, as illustrated in the third example.

Key takeaway: With independent Type I extreme-value shocks and unit scale, the log ratio of two conditional choice probabilities equals the difference in their choice-specific values. This is a state-by-state inversion of relative values, not a recovery of absolute value levels or current payoffs by itself.

Check Your Understanding

Use the normalized Type I extreme-value model unless a question specifies otherwise.

If $P_1(x)=0.25$ and $P_0(x)=0.75$, what is $v_1(x)-v_0(x)$? State the formula before calculating.
Why do conditional choice probabilities fail to identify the absolute level of every $v_j(x)$?
For three alternatives with probabilities $0.10$, $0.60$, and $0.30$, find the value differences for alternatives 1 and 2 relative to alternative 0.
If the shock scale is $\sigma=2$, how does the inversion formula change?
Why can a choice-specific value difference differ from the corresponding current payoff difference in a dynamic model?