<p>I would call this a mean-variance utility function. The agent likes higher mean values, which is the first term, but trades that off against higher variance, which is the second term.</p>

<p>If the random variable of interest is normally distributed with mean $P_{t+1} + \delta_{t+1} - (1+r^f)P_t$ and covariance matrix $\Omega$, and if the agent has constant absolute risk aversion utility, such as $u(w) = 1-e^{-\gamma w}$, then maximizing that utility is equivalent to maximizing mean-variance utility. <a href="http://www.tau.ac.il/~spiegel/teaching/corpfin/mean-variance.pdf" rel="nofollow">Here are some details I found with a quick google search</a>.</p>

CC BY-SA 3.0
2015-06-07T10:32:18.237Z
Hi Nick, thank you very much for you response. I still have one question though, more specifically would you consider this a quadratic, exponential or isoelastic utility function? As far as i remember mean-variance utility can be derived as either of the beforementioned? For instance you can derive CAPM which uses mean-variance utility, in either form isoelastic, quadratic, or expontential?
nvanlaer
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2015-06-07T13:01:27.920Z
Just to clarify wouldn't this be log utility function where $q{z}(R)=R-γ/2R^2$ ?
nvanlaer
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2015-06-07T17:05:16.760Z
@nvanlaer The specific function you have above is a quadratic function. When risk is normally distributed, then maximizing an exponential utility function (defined over *outcomes*, i.e. payoffs), is equivalent to maximizing a quadratic function (defined over level of exposure to the risk asset) But, for example, it won't be true that maximizing expected quadratic utility (again, defined over outcomes) or isoelastic utility will have the same property. I'm not sure if that answers your question because I'm not sure I understand your question. I don't know how to interpret your second comment.
NickJ
CC BY-SA 3.0
2015-05-25T12:27:33.710Z
2015-05-25T12:27:33.710Z
NickJ
<p>Can someone tell my how i can figure out which type of utility function the following maximization problem has. It is for an overlapping generations model.</p>

<p>$$\underset{x'}{\max} x'\big(E_t (P_{t+1}+δ_{t+1})-(1+r^f)P_t\big) - \tfrac{\gamma}{2} x'Ωx $$<br>
Where:
$x'=$ portfolio of shares </p>

<p>$E_t (P_{t+1}+δ_{t+1}) =$ expected future payoff </p>

<p>$\gamma =$ agent i risk aversion</p>

<p>$Ω=$ covariance matrix </p>

CC BY-SA 3.0
2015-05-24T21:31:21.530Z
Could you tidy up the notation? Lower script is _{what you want to have as lower index}.
E.g.: P_{t+1}
Giskard
<utility><expected-utility><olg>
How do I find out which form of utility function is being used?
nvanlaer
# How do I find out which form of utility function is being used?

(Asked by: username_0 on 2015-05-24)

Can someone tell my how i can figure out which type of utility function the following maximization problem has. It is for an overlapping generations model.

$$\underset{x'}{\max} x'\big(E\_t (P\_{t+1}+δ\_{t+1})-(1+r^f)P\_t\big) - \tfrac{\gamma}{2} x'Ωx $$  
Where:
$x'=$ portfolio of shares

$E\_t (P\_{t+1}+δ\_{t+1}) =$ expected future payoff

$\gamma =$ agent i risk aversion

$Ω=$ covariance matrix

**Tags:** utility, expected-utility, olg

> **Comments:**
> **username_1**: Could you tidy up the notation? Lower script is _{what you want to have as lower index}. E.g.: P_{t+1}  

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## 1 answers

### Answer by username_2 on 2015-05-25. Score: 2

I would call this a mean-variance utility function. The agent likes higher mean values, which is the first term, but trades that off against higher variance, which is the second term.

If the random variable of interest is normally distributed with mean $P\_{t+1} + \delta\_{t+1} - (1+r^f)P\_t$ and covariance matrix $\Omega$, and if the agent has constant absolute risk aversion utility, such as $u(w) = 1-e^{-\gamma w}$, then maximizing that utility is equivalent to maximizing mean-variance utility. [Here are some details I found with a quick google search](http://www.tau.ac.il/~spiegel/teaching/corpfin/mean-variance.pdf).

> **Comments:**
> **username_0**: Hi Nick, thank you very much for you response. I still have one question though, more specifically would you consider this a quadratic, exponential or isoelastic utility function? As far as i remember mean-variance utility can be derived as either of the beforementioned? For instance you can derive CAPM which uses mean-variance utility, in either form isoelastic, quadratic, or expontential?  
> **username_0**: Just to clarify wouldn't this be log utility function where $q{z}(R)=R-γ/2R^2$ ?  
> **username_2**: @username_0 The specific function you have above is a quadratic function. When risk is normally distributed, then maximizing an exponential utility function (defined over *outcomes*, i.e. payoffs), is equivalent to maximizing a quadratic function (defined over level of exposure to the risk asset) But, for example, it won't be true that maximizing expected quadratic utility (again, defined over outcomes) or isoelastic utility will have the same property. I'm not sure if that answers your question because I'm not sure I understand your question. I don't know how to interpret your second comment.  

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