# Understanding the two-stage choice paradigm

In Manski - The structure of random utility models the following example is proposed:

Consider the alternative set space a = $(\alpha,\beta,\gamma)$ with the attribute representation: $X = \begin{bmatrix} & \alpha & \beta & \gamma & \cr x_1 & 1 & 1 & 2 \cr x_2 & 0.5 & -0.5 & -1 \end{bmatrix}$.

The individuals/decision-makers $\sigma$ and $\tau$ are characterized by the attributes: $S = \begin{bmatrix} \sigma & \tau \cr 1 & 0 \end{bmatrix}$.

Finally the function $w(x,s)$ which defines the utility of the utility maximizing agents $\sigma$ and $\tau$ is given by $w(x,s) = x_1 + x_2\cdot s$.

The aim is to compute the probabilities: $P_t(a\in^c C)$ which is the probability that a decision-maker $t\in T$ will choose the alternative $a$ from a possible choice set $C$. By the definition of the alternative space a one can think of $C$ as any non-empty subset of any particular order of the alternatives $\alpha, \beta$ and $\gamma$. Hence there will be quite a few possible choice sets $C$ from which a decision-maker can choose. Manski denotes the choice set space, the set of all possible non-empty subsets $C$, with $\Gamma$ and $T$ represents the population of all possible decision-makers.

To calculate $P_t(a\in^c C)$ additional information is given. The joint distribution $M_{\Gamma T}((C,t))$ of possible choice sets $C$ and the decision-makers $t$ is given by:

$M_{\Gamma T}((\alpha,\beta,\gamma),\sigma) = \frac{2}{36}$ for each of the six ordered choice sets whose elements are $\alpha,\beta$ and $\gamma$

$M_{\Gamma T}((\alpha,\beta),\sigma) = \frac{1}{12}$ for each of the two ordered choice sets whose elements are $\alpha$ and $\beta$

$M_{\Gamma T}((\alpha,\beta,\gamma),\tau) = \frac{1}{36}$ for each of the six ordered choice sets whose elements are $\alpha,\beta$ and $\gamma$

$M_{\Gamma T}((\alpha,\beta),\tau) = \frac{2}{12}$ for each of the two ordered choice sets whose elements are $\alpha$ and $\beta$

To calculate the probabilities $P_t(a\in^c C)$ for the case that the information of $x_2$ is not observed we first need to calculate the vector $W_{Ct}$ which contains the utilities for the alternatives contained in $C$ for a decision-maker $t$. In this case this will yield:

$\bar W_{(\alpha,\beta,\gamma),\sigma} = \begin{pmatrix} 1.5\\ 0.5\\ 1\end{pmatrix}$ and $\bar W_{(\alpha,\beta,\gamma),\tau} = \begin{pmatrix} 1\\ 1\\ 2\end{pmatrix}$

For the calculation of the probability $P_t(a\in^c C)$ Manski gives the following two formulas:

$(1) \ P(\bar W_{Ct} | r_{Co},s_{to}) = \frac{ \sum_{ (\tilde C, \tilde t):r_{\tilde Co} = r_{Co}, s_{\tilde to} = s_{to}, W_{\tilde C\tilde t} = \bar W_{Ct} } M_{\Gamma T}((\tilde C,\tilde t))}{\sum_{ (\tilde C, \tilde t):r_{\tilde Co} = r_{Co}, s_{\tilde to} = s_{to}} M_{\Gamma T}((\tilde C,\tilde t))}$

$(2) \ P_t(a\in^c C) = \sum_{\bar W_{Ct}:\bar w_{at}\ge \bar w_{\tilde at}, \tilde a\in C}P(\bar W_{Ct} | r_{Co},s_{to})$

So clearly in order to calculate $P_t(a\in^c C)$ we need to calculate (1) first. This is done by summation over $M_{\Gamma T}((\tilde C,\tilde t))$ for all possible choice problems $(\tilde C,\tilde t)$ for particular values of $r_{\tilde Co},s_{\tilde to}$ and $W_{\tilde C\tilde t}$ namely $r_{Co},s_{to}$ and $\bar W_{Ct}$. To do that we need to consider the definition of $r_{Co}$ and $s_{to}$. Manski proposes that $X$ and $S$ are split up into observable and non-observable parts like $X = [X_o,X_u]$ and $S = [S_o,S_u]$ where the "o" donates the observable, and "u" the non observable part. Now we can write $X_o$ as $X_0 = (x_{ao}, \forall \ a\in$ a). The definition of $r_{Co}$ is now given by $r_{Co} = (x_{ao}, \forall \ a\in C)$. So the only difference between $X_o$ and $r_{co}$ is the set on which $x_{ao}$ is defined on, since $r_{Co}$ considers all possible choice-sets $C$, $X_o$ only defines $x_{ao}$ on the alternative space a.

Now my dilemma: Since I need to calculate (1) in order to calculate (2), I need to sum up the joint probabilities $M_{\Gamma T}((\tilde C, \tilde t))$ over possible values of $r_{Co}, s_{to}$ and $\bar W_{Ct}$. Since the value of $s_{to}$ is given by the integers $1$ or $0$ and $\bar W_{Ct}$ is just one possible utility-vector depending on which choice-set $C$ we are currently looking on, I only need to figure out what $r_{Co}$ actually is. The definition is quite clear but for this example I'm not quite sure. Since $x_2$ is not observed I thought of $r_{Co}$ as a vector of the observed outcomes of $x_1$, i.e., $r_{Co} = (1.5,0.5,1)$ for $C = (\alpha,\beta,\gamma)$. But if I consider $r_{Co}$ in this way the calculation will not work since (1) will always be 1 and the sum in (2) can get larger then 1 which is of course not possible.

I'm stuck with this problem a couple of days now and any help will be much appreciated!

This would be my answer for my question if someone would ask me. With no further knowledge, what the solotion would look like, though I already now the solution because it's part of the previous mentioned paper (page 237), I would argue like:

(it's going to be a quite long answer, but in my opinion necessary to explain all the details)

The solution, for the sets of three alternatives, i.e., for all ordered sets composed of $\alpha,\beta$ and $\gamma$ is given by Manski with:

$P_\sigma(\alpha\in^c (\alpha,\beta,\gamma))=\frac{1}{2}$

$P_\sigma(\beta\in^c (\alpha,\beta,\gamma))=\frac{1}{2}$

$P_\sigma(\gamma\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\alpha\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\beta\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\gamma\in^c (\alpha,\beta,\gamma))=1$

It's obvious that the sum $\sum_{a\in C}P_t(a\in^c C)$ needs to be $1$ for any $C$ and any $t\in T$.

If we calculate the utility-vectors $W_{Ct}$ for any $C$ and $t\in T$, i.e., for any ordered choice set $C$ composed of three alternatives $\alpha,\beta$ and $\gamma$ and any order choice set $C$ composed of the two alternatives $\alpha$ and $\beta$ we will get:

$\bar W_{(\alpha,\beta,\gamma),\sigma} = \begin{pmatrix} 1.5\\ 0.5\\ 1\end{pmatrix}$ for any combination of $(\alpha,\beta,\gamma)$ and $t=\sigma$

$\bar W_{(\alpha,\beta),\sigma} = \begin{pmatrix} 1.5\\ 0.5\end{pmatrix}$ for any combination of $(\alpha,\beta)$ and $t=\sigma$

$\bar W_{(\alpha,\beta,\gamma),\tau} = \begin{pmatrix} 1\\ 1\\ 2\end{pmatrix}$ for any combination of $(\alpha,\beta,\gamma)$ and $t=\tau$

$\bar W_{(\alpha,\beta),\tau} = \begin{pmatrix} 1\\ 1\end{pmatrix}$ for any combination of $(\alpha,\beta)$ and $t=\tau$

If we have a look at $t=\sigma$, with no further knowledge but the information that a decision-maker is a utility maximizing agent, we would think that in any case, i.e., for any choice set $C$ of three or two alternatives, the agent would always choose the alternative $\alpha$.

Similarly the agent $t=\tau$ would choose $\gamma$ in any case in which he picks a alternative from set composed of three alternatives. For the case in which he picks alternatives from a set composed of two alternatives he would be indifferent between $\alpha$ and $\beta$.

If we have a look at formula (1) these ideas would translate into:

$P(\bar W_{(\alpha,\beta,\gamma),\sigma} | r_{(\alpha,\beta,\gamma),o}, s_{\sigma,o})=\frac{\frac{2}{36}}{6\frac{2}{36}} = \frac{1}{6}$ for any of the six possible orderings of the vector $\bar W_{(\alpha,\gamma,\beta),\sigma}$, i.e., for each of the six ordered choice sets $C$ whose elements are $\alpha,\beta$ and $\gamma$.

$P(\bar W_{(\alpha,\beta),\sigma} | r_{(\alpha,\beta),o}, s_{\sigma,o})=\frac{\frac{1}{12}}{2\frac{1}{12}} = \frac{1}{2}$ for any of the two possible orderings of the vector $\bar W_{(\alpha,\gamma),\sigma}$, i.e., for each of the two ordered choice sets $C$ whose elements are $\alpha$ and $\beta$.

Similarly we get

$P(\bar W_{(\alpha,\beta,\gamma),\tau} | r_{(\alpha,\beta,\gamma),o}, s_{\tau,o})=\frac{2\frac{2}{36}}{6\frac{2}{36}} = \frac{1}{3}$

$P(\bar W_{(\alpha,\beta),\tau} | r_{(\alpha,\beta),o}, s_{\tau,o})=\frac{2\frac{1}{12}}{2\frac{1}{12}} = 1$

Comments: The reason for the $2$ being a part in the denominator of $P(\bar W_{(\alpha,\beta,\gamma),\tau} | r_{(\alpha,\beta,\gamma),o}, s_{\tau,o})$ and $P(\bar W_{(\alpha,\beta),\tau} | r_{(\alpha,\beta),o}, s_{\tau,o})$ is, that we only observe three distinct vectors of $W_{C,\tau}$, though we've six different orderings of the choice set $(\alpha,\beta,\gamma)$. For instance, for the two cases $C=(\alpha,\beta,\gamma)$ and $C'=(\beta,\alpha,\gamma)$, we will observe the same vector $W_{C,\tau}$. Since, by formula (1), we need to count how often we observe $W_{\tilde C, \tilde t} = \bar W_{C,t}$, this is what the $2$ in the denominator essentially does.

If we calculate (2), for a certain size of $C$ (like $|C|=3$), we need to consider all the values of $P(\bar W_{C,t} | r_{C,o}, s_{t,o})$ for which a value $\bar w_{a,t}$ is at least as big as all the other values $\bar w_{\tilde a, t}, \tilde a\in C$. If we do that, we will get the following values:

(all probabilities will be the same for any ordering of $(\alpha,\beta,\gamma)$. The "$\in^c (\alpha,\beta,\gamma)$" only adresses the set we are currently looking on)

$P_\sigma(\alpha\in^c (\alpha,\beta,\gamma))=6\frac{1}{6} = 1$

(there are six distinct cases of $W_{C,\sigma}$ in which the value $\bar w_{\alpha,\sigma}$ is greater then all the other values )

$P_\sigma(\beta\in^c (\alpha,\beta,\gamma))=0$

$P_\sigma(\gamma\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\alpha\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\beta\in^c (\alpha,\beta,\gamma))=0$

$P_\tau(\gamma\in^c (\alpha,\beta,\gamma))=3\frac{1}{3} = 1$

(there are only three distinct cases of $W_{C,\tau}$, hence the $3$ )

For $|C|=2$ we will get:

$P_\sigma(\alpha\in^c (\alpha,\beta))=2\frac{1}{2} = 1$

$P_\sigma(\beta\in^c (\alpha,\beta))=0$

Now the forumla (2) seems to fail for $t=\tau$, since the imposed condition not only accounts for the cases in which $\bar w_{a\tau}$ is greater then $w_{\tilde a,\tau}$ but also if the equality between those two quantities is true. In this particular case, $|C|=2,t=\tau$ all the values of $w_{\tilde a, \tau}, \tilde a\in C$ are equal. So by the definition of (2) we would count all the cases of $P(\bar W_{(\alpha,\beta),\tau} | r_{(\alpha,\beta),o}, s_{\tau,o})$ for each alternative. Here we only observe one possible value of $\bar W_{C,\tau}$, i.e., $W_{(\alpha,\beta),\tau}=W_{(\beta,\alpha),\tau}=\begin{pmatrix} 1\\ 1\end{pmatrix}$. So by (2) we would get:

$P_\sigma(\alpha\in^c (\alpha,\beta))=1$

$P_\sigma(\beta\in^c (\alpha,\beta))=1$

,since the condition "$\bar w_{a,\tau} \ge w_{\tilde a, \tau}, \tilde a\in C$" is true for both cases, i.e., for the summation of $a=\alpha$ and $a=\beta$. But this cannot be true because $\sum_{a\in (\alpha,\beta)}P_\tau(a\in^c (\alpha,\beta))=2\ne 1$. I would suggest the following work arround:

Instead of (2) I would suggest we could use

$(3) P_t(a\in^c C) = \sum_{\bar w_{a,t} > w_{\tilde a, t}, \tilde a\in C}P(\bar W_{C,t}|r_{Co},s_{to}) + \sum_{\bar w_{a,t} = w_{\tilde a, t}, \tilde a\in C}\frac{P(\bar W_{C,t}|r_{Co},s_{to})}{2}$

This equation would, in my optinion, better adress the issue of a agent being indifferent between all possible alternatives. The former mentioned issue, leaving all the other results unchanged, would be, by using (3) instead of (2):

$P_\sigma(\alpha\in^c (\alpha,\beta))=\frac{1}{2}$

$P_\sigma(\beta\in^c (\alpha,\beta))=\frac{1}{2}$