# How to derive formula for marginal probability of choosing nest in nested logit model?

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!

• Welcome to Economics SE. I don't understand all your notation, can you please explain what $W_{nk}, Y_{nj}, B_k$ are? Jul 23, 2015 at 21:38
• $W_{nk}$ 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. Jul 24, 2015 at 18:54

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 $\epsilon_{nk}$.