<p>Looking two slides before, the author provides another utility function, under "Separable Utility" this time:</p>

<p>$$U(c,n) = \ln(c) - a\frac {n^{1+1/v}}{1+1/v}$$</p>

<p>Comparing the two I believe we can safely reach the conclusion that the author had implicitly in mind "additive separability", i.e. a functional form where second partials are zero, irrespective of whether a transformation can separate the arguments of the function. At least in some corners, when one needs the logarithmic transformation to induce separability, one talks about "multiplicatively separable" preferences.</p>

CC BY-SA 3.0
2015-08-26T16:48:28.253Z
2015-08-26T16:48:28.253Z
Alecos Papadopoulos
<p>You are right, these preferences are separable. A way to see it is to notice that
\begin{equation*}
U(c,n) \geq U(c',n) \Rightarrow U(c,n') \geq U(c',n')
\end{equation*}
for any $n,n'$, and 
\begin{equation*}
U(c,n) \geq U(c,n') \Rightarrow U(c',n) \geq U(c',n')
\end{equation*}
for any $c,c'$.
They are even additively separable, as your log-transformation shows.</p>

CC BY-SA 3.0
2015-08-25T10:23:22.433Z
2015-08-25T10:29:15.353Z
2015-08-25T10:29:15.353Z
Oliv
<p>On <a href="http://www.econ.umn.edu/~vr0j/ec8503-10/daolufrisch.pdf" rel="nofollow">this presentation</a>, the last slide is titled "Non Separable Utility", and the preferences given are</p>

<p>$$ \frac{\left(c^\gamma (1-n)^{1-\gamma}\right)^{1-\sigma}}{1-\sigma}$$ </p>

<p>However, I can log-transform them as </p>

<p>$$ \log\left[ c^{\gamma(1-\sigma)} (1-n)^{(1-\gamma)(1-\sigma)} (1-\sigma)^{-1}\right]$$</p>

<p>which are clearly separable. Did I miss something?</p>

<utility>
Is this non-separable utility?
FooBar
# Is this non-separable utility?

(Asked by: username_0 on 2015-08-25)

On [this presentation](http://www.econ.umn.edu/~vr0j/ec8503-10/daolufrisch.pdf), the last slide is titled "Non Separable Utility", and the preferences given are

$$ \frac{\left(c^\gamma (1-n)^{1-\gamma}\right)^{1-\sigma}}{1-\sigma}$$

However, I can log-transform them as

$$ \log\left[ c^{\gamma(1-\sigma)} (1-n)^{(1-\gamma)(1-\sigma)} (1-\sigma)^{-1}\right]$$

which are clearly separable. Did I miss something?

**Tags:** utility


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## 2 answers

### Answer by username_2 on 2015-08-25. Score: 5 | Accepted answer

You are right, these preferences are separable. A way to see it is to notice that
\begin{equation\*}
U(c,n) \geq U(c',n) \Rightarrow U(c,n') \geq U(c',n')
\end{equation\*}
for any $n,n'$, and
\begin{equation\*}
U(c,n) \geq U(c,n') \Rightarrow U(c',n) \geq U(c',n')
\end{equation\*}
for any $c,c'$.
They are even additively separable, as your log-transformation shows.


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### Answer by username_1 on 2015-08-26. Score: 2

Looking two slides before, the author provides another utility function, under "Separable Utility" this time:

$$U(c,n) = \ln(c) - a\frac {n^{1+1/v}}{1+1/v}$$

Comparing the two I believe we can safely reach the conclusion that the author had implicitly in mind "additive separability", i.e. a functional form where second partials are zero, irrespective of whether a transformation can separate the arguments of the function. At least in some corners, when one needs the logarithmic transformation to induce separability, one talks about "multiplicatively separable" preferences.


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