<p>A utility function is additively separable if it can be written as:
$U(x,y) = u(x) + v(y)$</p>

<p>Examples:</p>

<p>*$U(x,y) =ax + by$ is additively separable by inspection.</p>

<p>*$U(x,y) = ax + bx2 + cy$ is also.</p>

<p>*$U(x,y) = x^a y^b$ is additively separable, because you can write it as  $U(x,y) = log(x^a)+log(y^b)=alog(x)+b log(y)= u(x) + v(y)$</p>

<p>*$U(x,y) = \frac{xy}{x+y}$ is not additively separable because there's no way to transform it into an independent sub-function of $x, y$. Even if you take logs, you're down to $U = log(x) + log(y) - log(x+y)$ - notice that the third term can't be 'split'. Generally, mixing addition, multiplication and exponentiation will destroy additive separability.</p>

<p>And so we can see, The actual definition of additive separability is: </p>

<p>A function $f(x_1,...,x_n)$ is AS if it can be rewritten as $f(x_1,...,x_n)=f_1(x_1)+...+f_n(x_n)$ </p>

<hr>

<p>The assumption is usually one of mathematical convenience. For example, when utility is additively separable in $x,y$, then the marginal utility of $x$ does not depend on the level of $y$, and vice-versa. And so anything requiring the use of partial derivatives is made easier. Is it a reasonable assumption? Sometimes. For example, if your utility depends on apples and horses, we can probably assume additive separability. If Instead your utility depends on two closely related things, it might be a bad assumption. </p>

<hr>

CC BY-SA 3.0
2017-04-06T12:48:37.803Z
a random down vote?
123
CC BY-SA 4.0
2021-12-08T07:22:43.117Z
Are you sure about $x^ay^b$? Your derivation considers its log, but not its original form. I do not think it is additively separable.
Richard Hardy
CC BY-SA 3.0
2017-04-01T15:20:09.003Z
2017-04-01T23:36:04.977Z
2017-04-01T23:36:04.977Z
123
<p>A function is additively separable in its arguments if it has the form</p>

<p>$$f(x,y) = g(x) + h(y)$$</p>

<p>This means that the cross partials are zero, and so there is no "cross" effect of the one argument over the <em>marginal</em> effect that the other has on the value of the function. Since marginal effects are at the very heart of Economics (<a href="https://economics.stackexchange.com/a/49/61">see here</a>), assuming additive separability greatly simplifies the analysis. In dynamic problems, where the <em>intertemporal</em> utility function is assumed to be additively separable, it permits us to transform an infinite horizon problem into a recursive two-period one.</p>

<p>Functions that can be transformed into something additively separable (by usually considering their logarithms), are sometimes called "multiplicatively separable". The most famous example here is the Cobb-Douglas production function:</p>

<p>$$Q = K^aL^{1-a} \implies \ln Q = a\ln K + (1-a)\ln L$$</p>

<p>As for linearity, it is a <em>unique</em> (structurally) relationship, while non-linear relationships are many, perhaps too many.
A mathematician once said that "the whole field of Analysis, is essentially the study of linear approximation of non-linear relations". </p>

<p>Again, mathematical tractability is the drive here, supported by the fact that a linearity assumption is a "first-order" approximation to the true relation (see Taylor expansion).</p>

CC BY-SA 3.0
2017-04-01T16:22:37.550Z
To complete this answer, it might be worth noting that the behavioral implications of additive separability are: \begin{equation} \text{ for all } (x,y,z,w), (x,y) \succ (z,y) \Leftrightarrow  (x,w) \succ (z,w) \end{equation}
and 
\begin{equation} \text{ for all } (x,y,z,w), (x,z) \succ (x,w) \Leftrightarrow  (y,z) \succ (y,w) \end{equation}
 This means that preferences over argument 1 are independent of the value of argument 2, and vice versa, which gives behavioral content to your point that there is no ``cross'' effect.
Oliv
CC BY-SA 3.0
2017-04-01T17:39:42.813Z
@Oliv This is a nice addition. I only mention that additive separability touches also relations that do not have to do with preferences (e.g. factors of production).
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-09T18:43:28.187Z
@AlecosPapadopoulos Alecos, I came across your response in explaining Method of Moments in stack exchange and wonder if I could ask few questions. Basically, I am trying to conceptually understand what Hayashi is keep mentioning in his Chapter 3 where he states something like "orthogonality conditions mean s set of population moments are all zero and MoM principle is to choose the parameter estimate so that corresponding sample moments are also equal to zero." I think I kind of understand this when he shows with equations, but don't really understand conceptually.
Frank Swanton
CC BY-SA 3.0
2017-04-09T18:44:56.390Z
@AlecosPapadopoulos In particular, because we start off with the classical linear regression model, then make a leap to relaxing those assumptions and jump to ergodic stationarity to deal with time series, I have a challenging time patching all these in a nice way... Would appreciate your help.
Frank Swanton
CC BY-SA 3.0
2017-04-09T18:48:06.360Z
@AlecosPapadopoulos Any good reference article regarding the concept of "matching population moments with sample moments" in the context of econometrics would be helpful. Not sure if Wooldrige or the like has something like that.
Frank Swanton
CC BY-SA 3.0
2017-04-09T18:48:51.870Z
@FrankSwanton It would be better to formulate this questions as a new Question here or better, over at CrossValidated, since these are statistical foundations rather than econometrics.
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-09T18:50:25.710Z
@AlecosPapadopoulos Yes, I was going to in CrossValidated, but I have 1 reputation. Let me try to post there, in the meantime any reference article recommendation would be great. Thanks.
Frank Swanton
CC BY-SA 3.0
2017-04-09T18:50:35.120Z
@FrankSwanton This is very fundamental: it has to do with applying the "principle of analogy", "as in population, so in sample". It is justified because we are looking for samples _representative_ of the population. If they are, then it must be the case that what holds at population level, should ~hold at sample level.
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-01T15:30:26.523Z
2017-04-01T17:56:13.830Z
2017-04-13T12:51:38.253Z
Alecos Papadopoulos
<p>Often I see both in micro and macro two common terminology :</p>

<ol>
<li><p>Additively separable. </p></li>
<li><p>Linear in price or linear in probability. </p></li>
</ol>

<p>I understand exactly as they sound  by looking at the functional form of the object. </p>

<p>But can someone provide why these structures or assumptions are sensible or unreasonable and why they are "convenient" or "useful"? The context can be anything consumer, producer, choice under uncertainty, game theory or GE. But trying to see why it is repeatedly coming up and why it is important or mathematically useful in many parts both in micro and macro. </p>

<macroeconomics><microeconomics><game-theory><consumer-theory><decision-theory>
Meaning of Additively Separable, Linear in X
Frank Swanton
# Meaning of Additively Separable, Linear in X

(Asked by: username_0 on 2017-04-01)

Often I see both in micro and macro two common terminology :

1. Additively separable.
2. Linear in price or linear in probability.

I understand exactly as they sound by looking at the functional form of the object.

But can someone provide why these structures or assumptions are sensible or unreasonable and why they are "convenient" or "useful"? The context can be anything consumer, producer, choice under uncertainty, game theory or GE. But trying to see why it is repeatedly coming up and why it is important or mathematically useful in many parts both in micro and macro.

**Tags:** macroeconomics, microeconomics, game-theory, consumer-theory, decision-theory


---

## 2 answers

### Answer by username_3 on 2017-04-01. Score: 8 | Accepted answer

A function is additively separable in its arguments if it has the form

$$f(x,y) = g(x) + h(y)$$

This means that the cross partials are zero, and so there is no "cross" effect of the one argument over the *marginal* effect that the other has on the value of the function. Since marginal effects are at the very heart of Economics ([see here](https://economics.stackexchange.com/a/49/61)), assuming additive separability greatly simplifies the analysis. In dynamic problems, where the *intertemporal* utility function is assumed to be additively separable, it permits us to transform an infinite horizon problem into a recursive two-period one.

Functions that can be transformed into something additively separable (by usually considering their logarithms), are sometimes called "multiplicatively separable". The most famous example here is the Cobb-Douglas production function:

$$Q = K^aL^{1-a} \implies \ln Q = a\ln K + (1-a)\ln L$$

As for linearity, it is a *unique* (structurally) relationship, while non-linear relationships are many, perhaps too many.
A mathematician once said that "the whole field of Analysis, is essentially the study of linear approximation of non-linear relations".

Again, mathematical tractability is the drive here, supported by the fact that a linearity assumption is a "first-order" approximation to the true relation (see Taylor expansion).

> **Comments:**
> **username_4**: To complete this answer, it might be worth noting that the behavioral implications of additive separability are: \begin{equation} \text{ for all } (x,y,z,w), (x,y) \succ (z,y) \Leftrightarrow  (x,w) \succ (z,w) \end{equation} and  \begin{equation} \text{ for all } (x,y,z,w), (x,z) \succ (x,w) \Leftrightarrow  (y,z) \succ (y,w) \end{equation}  This means that preferences over argument 1 are independent of the value of argument 2, and vice versa, which gives behavioral content to your point that there is no ``cross'' effect.  
> **username_3**: @username_4 This is a nice addition. I only mention that additive separability touches also relations that do not have to do with preferences (e.g. factors of production).  
> **username_0**: @AlecosPapadopoulos Alecos, I came across your response in explaining Method of Moments in stack exchange and wonder if I could ask few questions. Basically, I am trying to conceptually understand what Hayashi is keep mentioning in his Chapter 3 where he states something like "orthogonality conditions mean s set of population moments are all zero and MoM principle is to choose the parameter estimate so that corresponding sample moments are also equal to zero." I think I kind of understand this when he shows with equations, but don't really understand conceptually.  
> **username_0**: @AlecosPapadopoulos In particular, because we start off with the classical linear regression model, then make a leap to relaxing those assumptions and jump to ergodic stationarity to deal with time series, I have a challenging time patching all these in a nice way... Would appreciate your help.  
> **username_0**: @AlecosPapadopoulos Any good reference article regarding the concept of "matching population moments with sample moments" in the context of econometrics would be helpful. Not sure if Wooldrige or the like has something like that.  
> **username_3**: @FrankSwanton It would be better to formulate this questions as a new Question here or better, over at CrossValidated, since these are statistical foundations rather than econometrics.  
> **username_0**: @AlecosPapadopoulos Yes, I was going to in CrossValidated, but I have 1 reputation. Let me try to post there, in the meantime any reference article recommendation would be great. Thanks.  
> **username_3**: @FrankSwanton This is very fundamental: it has to do with applying the "principle of analogy", "as in population, so in sample". It is justified because we are looking for samples _representative_ of the population. If they are, then it must be the case that what holds at population level, should ~hold at sample level.  

---

### Answer by username_1 on 2017-04-01. Score: 8

A utility function is additively separable if it can be written as:
$U(x,y) = u(x) + v(y)$

Examples:

\*$U(x,y) =ax + by$ is additively separable by inspection.

\*$U(x,y) = ax + bx2 + cy$ is also.

\*$U(x,y) = x^a y^b$ is additively separable, because you can write it as $U(x,y) = log(x^a)+log(y^b)=alog(x)+b log(y)= u(x) + v(y)$

\*$U(x,y) = \frac{xy}{x+y}$ is not additively separable because there's no way to transform it into an independent sub-function of $x, y$. Even if you take logs, you're down to $U = log(x) + log(y) - log(x+y)$ - notice that the third term can't be 'split'. Generally, mixing addition, multiplication and exponentiation will destroy additive separability.

And so we can see, The actual definition of additive separability is:

A function $f(x\_1,...,x\_n)$ is AS if it can be rewritten as $f(x\_1,...,x\_n)=f\_1(x\_1)+...+f\_n(x\_n)$

---

The assumption is usually one of mathematical convenience. For example, when utility is additively separable in $x,y$, then the marginal utility of $x$ does not depend on the level of $y$, and vice-versa. And so anything requiring the use of partial derivatives is made easier. Is it a reasonable assumption? Sometimes. For example, if your utility depends on apples and horses, we can probably assume additive separability. If Instead your utility depends on two closely related things, it might be a bad assumption.

---

> **Comments:**
> **username_1**: a random down vote?  
> **username_2**: Are you sure about $x^ay^b$? Your derivation considers its log, but not its original form. I do not think it is additively separable.  

---

