# Estimating changes of choice probabilities with choice removal in nested logit

Suppose that I have a nested logit model. I can estimate the probability of making a particular choice for each decision maker. Now, I want to remove one or more of the choices (though each nest keeps at least one option). How can I estimate/simulate the new choice probabilities?

As a simple example, I might estimate a logit model (non-nested in this case, but you get the idea) over the choice of fruit. Consider an individual that actually chose a banana over an apple and an orange. I can estimate the probability that he would choose each kind of fruit. Suppose that a banana is no longer a choice. Now, what is the chance that he chooses the apple? And the orange?

Thanks.

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Are you saying you'd delete all observations who responded with the the removed choices? –  Macro Apr 13 '12 at 0:12
No, they are used for estimation, but then, when the model is used for prediction, those particular choices are not available. See the cheeky example that I offer in an edit. –  Charlie Apr 13 '12 at 1:41
So, is it right to interpret this as you wanting the conditional probability of each possible choice, given that you will not choose the removed categories? –  Macro Apr 13 '12 at 2:20
@Macro, yes, exactly. –  Charlie Apr 16 '12 at 14:01
In the basic logit example, you can think of removing an option as being equivalent to making that option so unattractive that nobody would ever choose it. Using your example, if the utility from the apple, orange, and banana are $u_a$, $u_o$, and $u_b$ respectively, then the estimated probability of choice $c=i$ is \begin{gather} p_i = \Pr[c=i|u_a,u_o,u_b]= \frac{\exp(u_i)}{\sum_j\exp(u_j)} \end{gather} If we make the banana less attractive, then its utility decreases. In the limit, it is not chosen, \begin{gather} \lim_{u_b\rightarrow-\infty}\Pr[c=b|u_a,u_o,u_b]=\frac{\exp(-\infty)}{\exp(-\infty)+\exp(u_a)+\exp(u_o)}=\frac{0}{0+\exp(u_a)+\exp(u_o)} \end{gather}
This limit is instructive: to calculate the marginal probability of the other choices, you simply remove the $\exp(u_b)$ term from the denominator. These probabilities thus increase by an amount proportional to their original choice probabilities. (Depending on your setting, this may or may not be a desirable feature. See IIA.) The probability of choosing option $i\neq b$ becomes \begin{gather} \Pr[c=i \neq b|u_a,u_o,u_b]=\frac{\exp(u_i)}{\exp(u_a)+\exp(u_o)}=\frac{\exp(u_a)+\exp(u_o)+\exp(u_b)}{\exp(u_a)+\exp(u_o)}p_i \end{gather}