<p>I think the answer depends on the space $Y$ is coming from. Consider if $Y \in \{ 1, \frac{T+1}{T}+\epsilon \}$, where $\epsilon &gt; 0$ and the probability $p$ of the better result is </p>

<p>$$p &gt; \frac{T^2}{\epsilon T + 1}$$</p>

<p>(Which just comes from solving:)</p>

<p>$$1\cdot (1-p) + \left(\frac{T+1}{T} + \epsilon \right)\cdot p &gt; T+1$$</p>

<p>Then no matter what utility function you have, risk averse or risk loving, you will always take the gamble (option B).</p>

<p>Did you mean to have some sort of restriction on $Y$ for this question? Otherwise we'd have to consider both $Y$ and $u$ to determine whether the player will take option B.</p>

CC BY-SA 3.0
2015-11-25T20:54:31.097Z
One restriction I had in mind is that the probability distribution of $Y$ does not depend on $T$ (i.e, the distribution is fixed, and based on it, you are allowed to select $T$).
Erel Segal-Halevi
CC BY-SA 3.0
2015-11-25T21:41:53.247Z
Oh, I wasn't meaning to have $Y$ based on $T$, just used it to set up an example of a distribution where all utility functions would pick such and such. Perhaps that's me making the example too convoluted. Being able to pick $T$ after knowing $Y$ though is something I never considered though.
Kitsune Cavalry
CC BY-SA 3.0
2015-11-27T06:15:33.957Z
OK, I edited the question to make it clear.
Erel Segal-Halevi
CC BY-SA 3.0
2015-11-25T19:49:56.540Z
2015-11-27T00:00:27.173Z
2015-11-27T00:00:27.173Z
Kitsune Cavalry
<p>Your utility from having $x$ dolars is $u(x)$. </p>

<p>There is a gamble in which the winnings in dollars are a random variable, $Y$. It is known that $E[u(Y)]&gt;E[u(1)]$, so you prefer to bet than to get one dollar for sure.</p>

<p>But now you are offered the following options. You select a large number $T$ (which may depend on the distribution of $Y$), and then you can choose between two options:</p>

<p>A. Receive $T+1$ dollars.</p>

<p>B. Bet $T$ times, where all bets are statistically independent and distributed like $Y$.</p>

<p>For what utility functions $u$ would you prefer option B (for a sufficiently large $T$)?</p>

<p>The expected utility of option B is:</p>

<p>$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right]$$</p>

<p>where $Y_t$ are i.i.d. variables distributed like $Y$. So the question is actually: for what functions is it true that, for a sufficiently large $T$:</p>

<p>$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right] &gt; u(1 + T)\; ?$$</p>

<p>One obvious answer is when $u(x)=x$, since in that case: </p>

<p>$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right] = \sum_{t=1}^T E[Y_t] = T E[Y_t]$$</p>

<p>since $E[Y_t]&gt;E[u(1)]=1$, it is clear that for a sufficiently large $T$:</p>

<p>$$T \cdot E[Y_t] &gt; T + 1$$</p>

<p>What other utility functions have this property?</p>

CC BY-SA 3.0
2015-11-25T18:25:06.763Z
any functions thats positive for x­­­>0 with T big enough would have that property.
Revoltic
CC BY-SA 3.0
2015-11-27T07:32:06.717Z
my intuition tells me that constant absolute risk aversion would work. with constant absolute risk aversion if I prefer a gamble over a fixed payoff at some income level, I will prefer it at all income levels. If the utility gain of the gamble is fixed across income levels, eventually for high enough $T$, the sequence of gambles will be preferred to $T+1$.
HRSE
CC BY-SA 3.0
2015-11-29T08:19:03.203Z
@HRSE I think you are right. What about other types of utility functions?
Erel Segal-Halevi
<utility><probability>
To bet or not to bet
Erel Segal-Halevi
# To bet or not to bet

(Asked by: username_0 on 2015-11-25)

Your utility from having $x$ dolars is $u(x)$.

There is a gamble in which the winnings in dollars are a random variable, $Y$. It is known that $E[u(Y)]>E[u(1)]$, so you prefer to bet than to get one dollar for sure.

But now you are offered the following options. You select a large number $T$ (which may depend on the distribution of $Y$), and then you can choose between two options:

A. Receive $T+1$ dollars.

B. Bet $T$ times, where all bets are statistically independent and distributed like $Y$.

For what utility functions $u$ would you prefer option B (for a sufficiently large $T$)?

The expected utility of option B is:

$$E\left[u\left(\sum\_{t=1}^T Y\_t\right)\right]$$

where $Y\_t$ are i.i.d. variables distributed like $Y$. So the question is actually: for what functions is it true that, for a sufficiently large $T$:

$$E\left[u\left(\sum\_{t=1}^T Y\_t\right)\right] > u(1 + T)\; ?$$

One obvious answer is when $u(x)=x$, since in that case:

$$E\left[u\left(\sum\_{t=1}^T Y\_t\right)\right] = \sum\_{t=1}^T E[Y\_t] = T E[Y\_t]$$

since $E[Y\_t]>E[u(1)]=1$, it is clear that for a sufficiently large $T$:

$$T \cdot E[Y\_t] > T + 1$$

What other utility functions have this property?

**Tags:** utility, probability
> **Comments:**
> **username_1**: any functions thats positive for x­­­>0 with T big enough would have that property.  
> **username_2**: my intuition tells me that constant absolute risk aversion would work. with constant absolute risk aversion if I prefer a gamble over a fixed payoff at some income level, I will prefer it at all income levels. If the utility gain of the gamble is fixed across income levels, eventually for high enough $T$, the sequence of gambles will be preferred to $T+1$.  
> **username_0**: @HRSE I think you are right. What about other types of utility functions?  

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## 1 answers

### Answer by username_3 on 2015-11-25. Score: 1

I think the answer depends on the space $Y$ is coming from. Consider if $Y \in \{ 1, \frac{T+1}{T}+\epsilon \}$, where $\epsilon > 0$ and the probability $p$ of the better result is

$$p > \frac{T^2}{\epsilon T + 1}$$

(Which just comes from solving:)

$$1\cdot (1-p) + \left(\frac{T+1}{T} + \epsilon \right)\cdot p > T+1$$

Then no matter what utility function you have, risk averse or risk loving, you will always take the gamble (option B).

Did you mean to have some sort of restriction on $Y$ for this question? Otherwise we'd have to consider both $Y$ and $u$ to determine whether the player will take option B.

> **Comments:**
> **username_0**: One restriction I had in mind is that the probability distribution of $Y$ does not depend on $T$ (i.e, the distribution is fixed, and based on it, you are allowed to select $T$).  
> **username_3**: Oh, I wasn't meaning to have $Y$ based on $T$, just used it to set up an example of a distribution where all utility functions would pick such and such. Perhaps that's me making the example too convoluted. Being able to pick $T$ after knowing $Y$ though is something I never considered though.  
> **username_0**: OK, I edited the question to make it clear.  

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