<p>The trade-off between risk and expected returns depends on your own preferences.</p>
<p>Assume that you are expected utility maximizer and let the return of the investment be given by the random variable <span class="math-container">$X$</span>. Your utility is given by.
<span class="math-container">$$
\mathbb{E}(u(X))
$$</span></p>
<p>Let <span class="math-container">$\mu$</span> be the mean of <span class="math-container">$X$</span> and let <span class="math-container">$\sigma^2$</span> be the variance of <span class="math-container">$X$</span> then taking a Taylor expansion of <span class="math-container">$u(x)$</span> around <span class="math-container">$u(\mu)$</span> gives:
<span class="math-container">$$
u(x) \approx u(\mu) + u'(\mu)(x - \mu) + \frac{u''(\mu)}{2}(x - \mu)^2
$$</span>
Taking expectations on both sides gives:
<span class="math-container">$$
\begin{align*}
\mathbb{E}(u(X)) &amp;\approx u(\mu) + u'(\mu)(\mu - \mu) + \frac{u''(\mu)}{2}\sigma^2,\\
&amp;= u(\mu) + \frac{u''(\mu)}{2} \sigma^2.\\
\end{align*}
$$</span></p>
<p>So the higher the curvature of <span class="math-container">$u$</span> (the more negative <span class="math-container">$u''(\mu)$</span>) the more negative the second term will be. Intuitively, the curvature measures the degree of aversion for uncertainty.</p>
<p>Remark that this approximation will only be good when <span class="math-container">$X$</span> does not deviate too much from the mean, so the Taylor expansion is good.</p>
<p>You can also take a Taylor approximation around zero. This gives:
<span class="math-container">$$
\begin{align*}
u(x) \approx u(0) + u'(0) x + \frac{u''(0)}{2} x^2,\\
\end{align*}
$$</span>
So:
<span class="math-container">$$
\mathbb{E}(u(X)) \approx u(0) + u'(0) \mu + \frac{u''(0)}{2} (\sigma^2 + \mu^2)
$$</span></p>
<p>See also the paper of <a href="https://www.jstor.org/stable/1807366" rel="nofollow noreferrer">Levy and Markowitz (1979), &quot;Approximating Expected Utility by a Function of Mean and Variance&quot;, American economic review, 69, 308-317</a>.</p>

CC BY-SA 4.0
2021-10-05T08:42:32.377Z
2021-10-05T08:42:32.377Z
tdm
<p>Supposing I can choose to invest money in several different investments, each having</p>
<ul>
<li>risk <span class="math-container">$\sigma_i$</span>, for example, calculated as standard deviation</li>
<li>and expected return <span class="math-container">$r_i$</span></li>
<li>let's assume they have the same duration and same initial investment to keep things simpler</li>
</ul>
<p>For example if <span class="math-container">$\sigma_1 = \sigma_2$</span>, I can easily choose the one for which <span class="math-container">$r_i$</span> is greater. if for example <span class="math-container">$\sigma_1 = 1.1\sigma_2$</span> and <span class="math-container">$r_1 = 2r_2$</span>, reasonably <span class="math-container">$r_1$</span> is better.
From the above, I could decide to pick the investment for which <span class="math-container">$\frac{r_i}{\sigma_i}$</span> is greater. But this is just empirical and arbitrary. Is this method any good? Is there any more rigorous way of choosing between investments?</p>

<investment><probability>
How to compare investments with different risk and expected return?
Paolo
# How to compare investments with different risk and expected return?

(Asked by: username_0 on 2021-10-04)

Supposing I can choose to invest money in several different investments, each having

* risk $\sigma\_i$, for example, calculated as standard deviation
* and expected return $r\_i$
* let's assume they have the same duration and same initial investment to keep things simpler

For example if $\sigma\_1 = \sigma\_2$, I can easily choose the one for which $r\_i$ is greater. if for example $\sigma\_1 = 1.1\sigma\_2$ and $r\_1 = 2r\_2$, reasonably $r\_1$ is better.
From the above, I could decide to pick the investment for which $\frac{r\_i}{\sigma\_i}$ is greater. But this is just empirical and arbitrary. Is this method any good? Is there any more rigorous way of choosing between investments?

**Tags:** investment, probability


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## 1 answers

### Answer by username_1 on 2021-10-05. Score: 4 | Accepted answer

The trade-off between risk and expected returns depends on your own preferences.

Assume that you are expected utility maximizer and let the return of the investment be given by the random variable $X$. Your utility is given by.
$$
\mathbb{E}(u(X))
$$

Let $\mu$ be the mean of $X$ and let $\sigma^2$ be the variance of $X$ then taking a Taylor expansion of $u(x)$ around $u(\mu)$ gives:
$$
u(x) \approx u(\mu) + u'(\mu)(x - \mu) + \frac{u''(\mu)}{2}(x - \mu)^2
$$
Taking expectations on both sides gives:
$$
\begin{align\*}
\mathbb{E}(u(X)) &\approx u(\mu) + u'(\mu)(\mu - \mu) + \frac{u''(\mu)}{2}\sigma^2,\\
&= u(\mu) + \frac{u''(\mu)}{2} \sigma^2.\\
\end{align\*}
$$

So the higher the curvature of $u$ (the more negative $u''(\mu)$) the more negative the second term will be. Intuitively, the curvature measures the degree of aversion for uncertainty.

Remark that this approximation will only be good when $X$ does not deviate too much from the mean, so the Taylor expansion is good.

You can also take a Taylor approximation around zero. This gives:
$$
\begin{align\*}
u(x) \approx u(0) + u'(0) x + \frac{u''(0)}{2} x^2,\\
\end{align\*}
$$
So:
$$
\mathbb{E}(u(X)) \approx u(0) + u'(0) \mu + \frac{u''(0)}{2} (\sigma^2 + \mu^2)
$$

See also the paper of [Levy and Markowitz (1979), "Approximating Expected Utility by a Function of Mean and Variance", American economic review, 69, 308-317](https://www.jstor.org/stable/1807366).


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