Understanding the two-stage choice paradigm

Question

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!

Answer 1

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):

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

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