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

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

<p>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
$$P_{nB_k} = P\left[(\epsilon_{nk}+\xi_{nk})-(\epsilon_{nk}+\xi_{nk}) \geq (W_{nk}+\lambda_kI_{nk})-(W_{ns}+\lambda_sI_{ns}), \forall s\neq k \right]\\
=\frac{e^{W_{nk}+\lambda_kI_{nk}}}{\sum_{s=1}^Ke^{W_{ns}+\lambda_sI_{ns}}} $$</p>

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

CC BY-SA 3.0
2015-07-26T02:07:35.850Z
2015-07-26T02:07:35.850Z
Daria
<p>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. </p>

<p>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</p>

<p>$$ U_{njk} = W_{nk} + Y_{nj} + \epsilon_{nk} + e_{nj} $$</p>

<p>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. </p>

<p>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</p>

<p>$$ EU_{nk} = W_{nk} + \lambda_kI_{nk} + \epsilon_{nk}$$</p>

<p>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>

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

<p>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. </p>

<p>Thank you!</p>

CC BY-SA 3.0
2015-07-23T21:38:06.427Z
Welcome to Economics SE. I don't understand all your notation, can you please explain what $W_{nk}, Y_{nj}, B_k$ are?
Giskard
CC BY-SA 3.0
2015-07-24T18:54:11.753Z
$W_{nk}$ 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.
Daria
<academic-graduate><decision-theory><choice-theory>
How to derive formula for marginal probability of choosing nest in nested logit model?
Daria
# How to derive formula for marginal probability of choosing nest in nested logit model?

(Asked by: username_0 on 2015-07-23)

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!

**Tags:** academic-graduate, decision-theory, choice-theory

> **Comments:**
> **username_1**: Welcome to Economics SE. I don't understand all your notation, can you please explain what $W_{nk}, Y_{nj}, B_k$ are?  
> **username_0**: $W_{nk}$ 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.  

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## 1 answers

### Answer by username_0 on 2015-07-26. Score: 1 | Accepted answer

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 $\epsilon\_{nk}$.


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