<p>For a risk lover, r is negative. Making the term $-\frac{r}{2}Var(X)$ positive. Thus, as variance increases, $E(x)-\frac{r}{2}Var(X)$ increases, which means their utility increases.</p>

CC BY-SA 3.0
2018-04-04T14:44:51.920Z
2018-04-04T14:44:51.920Z
Jerry
<p>Let $X$ denote wealth. The mean variance utility for a risk-averse person is given by $E(X)-\frac{r}{2}Var(X)$ where $r$ is degree of risk-version. $r=0$ implies that person is risk-neutral. How does the mean variance utility function look like for a risk-lover?</p>

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2018-04-03T07:50:15.163Z
He/She is risky neutral...not a risk lover!
user17871
CC BY-SA 3.0
2018-04-03T11:57:05.900Z
Well then it sounds like you think you have your answer? Are you just looking for confirmation? If so state that in the question and put your calculations in there too.
BB King
CC BY-SA 3.0
2015-03-24T10:37:15.783Z
This seems like a homework problem. Please show what you tried thus far or this is likely to be closed.
BKay
CC BY-SA 3.0
2015-03-24T16:41:04.773Z
Derived and could show that, utility function for risk loving guy is $E(X)+\frac{r}{2}Var(X);r\geq 0$.
Pradipta
<utility>
mean variance utility function for risk loving person
Pradipta
# mean variance utility function for risk loving person

(Asked by: username_0 on 2015-03-24)

Let $X$ denote wealth. The mean variance utility for a risk-averse person is given by $E(X)-\frac{r}{2}Var(X)$ where $r$ is degree of risk-version. $r=0$ implies that person is risk-neutral. How does the mean variance utility function look like for a risk-lover?

**Tags:** utility
> **Comments:**
> **username_3**: This seems like a homework problem. Please show what you tried thus far or this is likely to be closed.  
> **username_0**: Derived and could show that, utility function for risk loving guy is $E(X)+\frac{r}{2}Var(X);r\geq 0$.  
> **username_1**: He/She is risky neutral...not a risk lover!  
> **username_2**: Well then it sounds like you think you have your answer? Are you just looking for confirmation? If so state that in the question and put your calculations in there too.  

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## 1 answers

### Answer by username_4 on 2018-04-04. Score: 1

For a risk lover, r is negative. Making the term $-\frac{r}{2}Var(X)$ positive. Thus, as variance increases, $E(x)-\frac{r}{2}Var(X)$ increases, which means their utility increases.


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