<p>Normally, decreasing MRS means that the MRS decreases as one moves <strong>along</strong> the indifference curve in the direction of <span class="math-container">$x$</span>. As you compute it, you increase <span class="math-container">$x$</span> but you do not move along the indifference curve because you do not keep the utility level constant.</p>
<p>Consider the utility level <span class="math-container">$U$</span>. Then the equation for the indifference curve with level <span class="math-container">$U$</span> is:
<span class="math-container">$$
x_1 + x_2^\alpha = U \to x_2 = (U - x_1)^{1/\alpha}
$$</span>
(note that we only evaluate the right hand side over <span class="math-container">$x_1 \le U$</span> for <span class="math-container">$x_2$</span> not to become negative).</p>
<p>The MRS is then:
<span class="math-container">$$
MRS = \left|\frac{dx_2}{dx_1}\right| = \left|-\frac{1}{\alpha}(U - x_1)^{\frac{1 - \alpha}{\alpha}}\right| = \frac{1}{\alpha}(U - x_1)^{\frac{1 - \alpha}{\alpha}}.
$$</span>
Now, if you evaluate this at <span class="math-container">$U = x_1 + x_2^\alpha$</span>, you indeed obtain that:
<span class="math-container">$$
\left|\frac{dx_2}{dx_1}\right|_{U = x_1 + x_2^\alpha} = \frac{1}{\alpha x_2^{\alpha - 1}}.
$$</span>
However, the change in the MRS due to a change in <span class="math-container">$x_1$</span> along the indifference curve with utility value <span class="math-container">$U$</span> is:
<span class="math-container">$$
\frac{1}{dx_1}\left|\frac{d x_2}{d x_1}\right| = -\frac{(1 - \alpha)}{\alpha^2}(U - x_1)^{\frac{1 - 2 \alpha}{\alpha}}.
$$</span>
which is negative for <span class="math-container">$\alpha &lt; 1$</span> (note that <span class="math-container">$x_1 \le U$</span>).</p>

CC BY-SA 4.0
2024-09-23T14:52:39.033Z
2024-09-23T14:52:39.033Z
tdm
<p>Let the utility function be: <span class="math-container">$U(x_1,x_2) = x_1 + x_2^\alpha$</span>.
Diminishing MRS requires <span class="math-container">$\frac{dMRS}{dx_1} &lt;0$</span>, however, taking this derivative results in 0, as <span class="math-container">$MRS = \frac{1}{\alpha x_2^{\alpha -1}}$</span>.</p>
<ol>
<li><p>How do we interpret this result?</p>
</li>
<li><p>Does diminishing MRS hold when the second order condition of <span class="math-container">$x_1$</span> is 0?</p>
</li>
<li><p><span class="math-container">$\frac{dMRS}{dx_2} = \frac{1-\alpha}{\alpha x_2^\alpha} &gt; 0$</span>, when <span class="math-container">$\alpha &lt; 1$</span>, is this enough to confirm diminishing MRS?</p>
</li>
<li><p>Can <span class="math-container">$\alpha$</span> be negative? Does that make sense in a utility function?</p>
</li>
</ol>

<utility><consumer-theory><optimization>
Under what conditions would a quasilinear utility function in a function form exhibit diminishing marginal rate of substitution?
Mathemafrick
# Under what conditions would a quasilinear utility function in a function form exhibit diminishing marginal rate of substitution?

(Asked by: username_0 on 2024-09-21)

Let the utility function be: $U(x\_1,x\_2) = x\_1 + x\_2^\alpha$.
Diminishing MRS requires $\frac{dMRS}{dx\_1} <0$, however, taking this derivative results in 0, as $MRS = \frac{1}{\alpha x\_2^{\alpha -1}}$.

1. How do we interpret this result?
2. Does diminishing MRS hold when the second order condition of $x\_1$ is 0?
3. $\frac{dMRS}{dx\_2} = \frac{1-\alpha}{\alpha x\_2^\alpha} > 0$, when $\alpha < 1$, is this enough to confirm diminishing MRS?
4. Can $\alpha$ be negative? Does that make sense in a utility function?

**Tags:** utility, consumer-theory, optimization


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## 1 answers

### Answer by username_1 on 2024-09-23. Score: 1

Normally, decreasing MRS means that the MRS decreases as one moves **along** the indifference curve in the direction of $x$. As you compute it, you increase $x$ but you do not move along the indifference curve because you do not keep the utility level constant.

Consider the utility level $U$. Then the equation for the indifference curve with level $U$ is:
$$
x\_1 + x\_2^\alpha = U \to x\_2 = (U - x\_1)^{1/\alpha}
$$
(note that we only evaluate the right hand side over $x\_1 \le U$ for $x\_2$ not to become negative).

The MRS is then:
$$
MRS = \left|\frac{dx\_2}{dx\_1}\right| = \left|-\frac{1}{\alpha}(U - x\_1)^{\frac{1 - \alpha}{\alpha}}\right| = \frac{1}{\alpha}(U - x\_1)^{\frac{1 - \alpha}{\alpha}}.
$$
Now, if you evaluate this at $U = x\_1 + x\_2^\alpha$, you indeed obtain that:
$$
\left|\frac{dx\_2}{dx\_1}\right|\_{U = x\_1 + x\_2^\alpha} = \frac{1}{\alpha x\_2^{\alpha - 1}}.
$$
However, the change in the MRS due to a change in $x\_1$ along the indifference curve with utility value $U$ is:
$$
\frac{1}{dx\_1}\left|\frac{d x\_2}{d x\_1}\right| = -\frac{(1 - \alpha)}{\alpha^2}(U - x\_1)^{\frac{1 - 2 \alpha}{\alpha}}.
$$
which is negative for $\alpha < 1$ (note that $x\_1 \le U$).


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