Is this a case of nonseparable utility (across states of nature)?

Question

There are two states of nature: summer (hot) and winter (cold).
I have a utility function indexed by states of nature: $u(\cdot;summer)$ and $u(\cdot;winter)$.
There are two good to choose between: ice tea and hot tea.
In the summer it is hot and I prefer ice tea: $u(ice \ tea;summer)>u(hot \ tea;summer)$.
In the winter it is cold and I prefer hot tea: $u(ice \ tea;winter)<u(hot \ tea;winter)$.

Is that a valid example of a utility function that is nonseparable across states of nature?

Edit: summer vs. winter may imply different times, yet the example was meant to imply different states of nature while holding the time fixed. That was a poorly thought through choice on my end; sorry. Replace summer with shine and winter with rain for a less confusing example. (I am not doing that above to keep the already posted answer and comments relevant.)

Answer 1

No, I don't think so. Let $a_i$ be your choice (e.g. ice tea) in situation $i$ (e.g. summer).

Separability over states says that if for some $b_j$, $a_i$ and $c_i$: $$ u(a_i, b_j) \ge u(c_i, b_j), $$ then for all $d_j$ $$ u(a_i, d_j) \ge u(c_i, d_j). $$ Notice that both comparisons keep the choice in some state fixed but this choice changes. So your preference of $a$ over $c$ in state $i$ does not depend on what you have in state $j$.

So for example, if you prefer ice tea in summer over hot tea in summer when you have ice tea in the winter, then you should also prefer ice tea in summer over hot tea in the summer when you have hot tea in the winter.

A violation of separability would be that your preferences of ice tea versus hot tea in summer depends on what you have in the winter. For example, you prefer ice tea over hot tea in summer when you have hot tea in winter, but you prefer hot tea over ice tea in summer when you have ice tea in the winter.

Answer 2

I think the core issue with this question (and the other related one the OP posted Nonseparable utility across states of nature: an intuitive example) is we need to clarify what is meant by "separable".

Unfortunately, "separable" is among the most overused adjectives across formal theories, in econ and beyond (including in pure math itself, see https://en.wikipedia.org/wiki/Separability). It's also commonly used in casual speech about formal models as an informal allusion to some sort of invariance (the video linked in the related question is a good example).

So it's impossible to make progress on any question about "separability" without defining more formally what "separable" is supposed to mean, which I believe will show that the answer to the OP's question is unavoidably definition-specific.

Fundamentals (keeping things simple here, for illustration purposes only)

• Two states: $S = \{s, s'\}$.
• Set of outcomes $X$.
• Preference represented by $u$ over the set of all $[(s,a), (s',b) | s'']$, with $a,b \in X$ and $s''$ representing the state you are currently in (either the state that materialized, if only one state can, or the current state of nature if states represent successive events).

Within-state Separability

(The name is made-up. I don't claim this is standard terminology or a good choice thereof. There are also many variants of the property you could think of, e.g., by holding one of the two outcomes fixed in the "other" state.)

$$ u[(s,a), (s',c)|s] \geq u[(s,b), (s',d)|s] \text{ for some $c,d \in X$}$$

implies

$$ u[(s,a), (s', e)|s] \geq u[(s,b), (s',f)|s] \text{ for all $e,f \in X$},$$

and

$$ u[(s,c), (s',a)|s'] \geq u[(s,d), (s',b)|s'] \text{ for some $c,d \in X$}$$

implies

$$ u[(s,e), (s', a)|s'] \geq u[(s,f), (s',b)|s'] \text{ for all $e,f \in X$}$$

In words, Within-state Separability says that conditional on being in state $s^*$, your preferences over what happens in that state are independent of what [could have happened/has happened/will happen] in the other state.

However, Within-state Separability allows preferences to be "state-dependent" in the sense of having $u[(s,a), (s',c)|s] > u[(s,b), (s',d)|s]$ but $u[(s,c), (s',a)|s'] < u[(s,d), (s',b)|s']$.

In this sense (which seems to be close to what @tdm has in mind), the example your provide fails to be non-separable. As @tdm suggests, with this definition of separability, a non-separable preference would require something like $u[(s,a), (s',c)|s] > u[(s,b), (s',c)|s]$ but $u[(s,a), (s',d)|s] < u[(s,b), (s',d)|s]$, i.e., the outcome you [had/will have/could have had] in state $s'$ impacts the way you rank outcomes in state $s$ (conditional on being in state $s$).

Between-state Separability

(Similar warning applies)

$$ u[(s,a), (s',c)|s] \geq u[(s,b), (s',d)|s] \text{ for some $c,d \in X$}$$

if and only if

$$ u[(s,c), (s', a)|s'] \geq u[(s,d), (s',b)|s']$$

In words, Between-state Separability says that, conditional on being in state $s$, you prefer getting outcome $a$ over $b$ in state $s$ if and only if you would also prefer $a$ over $b$ in state $s'$ conditional on being in that state (and provided what you [had/will have/could have had] in the other state is held fixed).

However, Between-state Separability allows preferences in one state to depend on what you get in the other state in the sense of having $u[(s,a), (s',c)|s] > u[(s,b), (s',d)|s]$ but $u[(s,a), (s',e)|s] < u[(s,b), (s',f)|s]$.

If separability is understood as Between-state Separability (which seems to be closer to the definition of separability you have in mind), then the example you suggested is non-separable.