3
$\begingroup$

This may be an obvious question, but I'm flummoxed.

Let's say I've got a nested logit model with a degenerate nest—say, a healthcare choice model where the branches are Home vs. Hospital. The Home branch has only one twig, called NoTreatment, while the Hospital branch has six twigs, representing each of the possible hospitals.

I have data from a study showing coefficients on each of the alternative and individual characteristics within the Hospital branch, as well as the inclusive value, so I can easily construct

$$U_{nj|k} = \beta'X_{j|k} + \varepsilon_{j|k}$$

where $n$ represents the individual, $j$ represents the hospital, and $k$ represents the branch (ie, that the individual chose to be treated at a hospital in the first place).

My understanding is that the probability that individual $n$ chooses provider $j$ is:

$$P_{nj} = \frac{e^{\frac{V_j}{\sigma}}\left[\sum_{m\ne j}e^{\frac{V_m}{\sigma}}\right]^{\sigma -1}}{\left[\sum_{m}e^{\frac{V_m}{\sigma}}\right]^{\sigma }}$$

where $\sigma-1$ is the inclusive value.

Assuming that $U_{home} = 0+\varepsilon$, what is the probability that someone chooses to be home? Is it simply:

$$P_{n0} = \frac{\left[\sum_{m\ne home}e^{\frac{V_m}{\sigma}}\right]^{\sigma -1}}{\left[\sum_{m}e^{\frac{V_m}{\sigma}}\right]^{\sigma }}$$

or do I need to deal with $\sigma$ differently?

EDIT: grammar

$\endgroup$
0
$\begingroup$

I'm updating this to put the correct answer in. According to Train (see esp. equations 4.4 and 4.5), this should be calculated in two levels. The first (upper) level is the probability of the respondent picking the degenerate branch vs. the hospital branch.

Let $W_{nk}$ represent the deterministic portion of the decision-maker's utility for branch-specific characteristics (maybe average costs of the hospitals, average distances, etc). If $\lambda_i$ is the coefficient on the inclusive value for any non-degenerate branch, then the probability of individual $n$ choosing a branch $B_k$ is

$$pB_{nk}=\frac{e^{W_{nk}+\lambda_{k}I_{nk}}}{\sum_{m=1}^{K}e^{W_{nm}+\lambda_{m}I_{nm}}}$$

Because we assume that $W_{nk}=0$ when $k$ is the Home branch, and because we only have two branches, then

$$pB_{home}=\frac{1}{1+e^{W_{nm}+\lambda_{m}I_{nm}}}$$

where $B_m$ is the hospital branch.

The inclusive value, $I_{nm}$ comes from the lower-level probability estimation equation within the hospital branch. That is:

$$I_{nm}=\ln\sum_{j\in B_m}e^{V_{nj}/\lambda_m}$$

where $V_{nj}$ is the deterministic portion of the decision-maker's utility for each individual hospital within the hospital branch (as compared to $W$ which is the utility for the overall branch itself)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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