I am trying to understand all the details of the nested logit and what confuses me is the formula for marginal probability of choosing the nest. In more details: the joint probability of individual n choosing alternative j can be factored as the probability of individual n choosing nest k, multiplied by the probability of individual i choosing j conditional on having chosen nest k.

As I understand we are decomposing the decision process into two models: upper and lower. In the upper model decision maker chooses a nest and in the lower - alternative within the nest. Say utility of individual n choosing alternative j in the nest k is

$$ U_{njk} = W_{nk} + Y_{nj} + \epsilon_{nk} + e_{nj} $$

where $e_{nj}$ is EV I with scale parameter $\lambda_k$ and $\epsilon_{nk}$ is such that composite error term is EV I with scale parameter 1.

The lower model is trivial, it is a simple logit. However, the upper model is not clear for me. The expected utility of individual n from choosing nest k is

$$ EU_{nk} = W_{nk} + \lambda_kI_{nk} + \epsilon_{nk}$$

where $\lambda_kI_{nk}$ is expected utility that n will get from choice within the nest. And the marginal probability of choosing nest k is

$$P_{nB_k}=\frac{e^{W_{nk}+\lambda_kI_{nk}}}{\sum_{l=1}^K e^{W_{nl}+\lambda_kI_{n}}}$$

My question is how does this probability have a logit form if $\epsilon_{nk}$ is not extreme value? Or is it? Because as far as I understand the sum of two extreme value variables is not an extreme value.

Thank you!

  • 1
    $\begingroup$ Welcome to Economics SE. I don't understand all your notation, can you please explain what $W_{nk}, Y_{nj}, B_k$ are? $\endgroup$
    – Giskard
    Jul 23, 2015 at 21:38
  • $\begingroup$ $W_{nk}$ and $Y_{nj}$ are deterministic parts of the utility. $W_{nk}$ varies only between nests and is constant within nest, say $W_{nk} = \beta X_{nk}$, were $X_{nk}$ are characteristics of the nest. Same for $Y_{nj}$, but it varies within the nest as well. $\endgroup$
    – Daria
    Jul 24, 2015 at 18:54

1 Answer 1


As it turned out, my previous logic was wrong. Here is how it should be done.

Marginal probability of choosing nest $k$ is $$P_{nB_k} = P\left[\max_{j\in B_k} U_{njk} \geq \max_{j\in B_s} U_{njs}, \forall s\neq k \right]\\ = P\left[W_{nk}+\epsilon_{nk}+\max_{j\in B_k}(Y_{nj}+e_{nj}) \geq W_{ns}+\epsilon_{ns}+\max_{j\in B_s}(Y_{nj}+e_{nj}), \forall s\neq k \right]$$

Then as $e_{nj}$ is iid $Gumbel(0,\lambda_t)$, $\max_{j\in B_k}(Y_{nj}+e_{nj})$ is iid Gumbel with location parameter $\lambda_kI_{nk}$ and scale parameter $\lambda_k$. The Gumbel distribution is preserved over linear transformations so $$ \max_{j\in B_k}(Y_{nj}+e_{nj})=\lambda_kI_{nk}+\xi_{nk}$$ where $\xi_{nk}$ is iid $Gumbel(0,\lambda_t)$. Substituting back to marginal probability, get $$P_{nB_k} = P\left[(\epsilon_{nk}+\xi_{nk})-(\epsilon_{nk}+\xi_{nk}) \geq (W_{nk}+\lambda_kI_{nk})-(W_{ns}+\lambda_sI_{ns}), \forall s\neq k \right]\\ =\frac{e^{W_{nk}+\lambda_kI_{nk}}}{\sum_{s=1}^Ke^{W_{ns}+\lambda_sI_{ns}}} $$

the last equality follows from the fact that $\epsilon_{nk}+\xi_{nk}$ is iid $Gumbel(0,1)$ by assumption on $\epsilon_{nk}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.