<blockquote>
  <p>The intuitive way</p>
</blockquote>

<p>Draw the function <span class="math-container">$L_t$</span>, the graph below is build with <span class="math-container">$\gamma = 0.8$</span> and <span class="math-container">$\pi_t^\otimes = 1$</span></p>

<p><a href="https://i.sstatic.net/MC0qI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MC0qI.png" alt="enter image description here"></a></p>

<p>Convex in this context means that the function <span class="math-container">$L_t$</span> looks like a bowl, just one well defined minimum. And if you cut it with a horizontal plane, the result is an ellipse.</p>

<p><a href="https://i.sstatic.net/ISzGm.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ISzGm.gif" alt="enter image description here"></a></p>

<blockquote>
  <p>The formal way</p>
</blockquote>

<p>Just for convenience, define the variable <span class="math-container">$p_t = \pi_t - \pi_t^\otimes$</span> so the function becomes</p>

<p><span class="math-container">$$
L_t(p_t, Y_t) = \gamma p_t^2 + Y_t^2 \tag{B1}
$$</span></p>

<p>First note that if <span class="math-container">$\gamma &gt; 0$</span>  then <span class="math-container">$L_t(p_t, Y_t) \ge 0$</span>. To prove convexity you just need to show that for any pair <span class="math-container">$(p_{t,1},Y_{t,1})$</span> and <span class="math-container">$(p_{t,2},Y_{t,2})$</span> the condition</p>

<p><span class="math-container">$$
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &lt; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B2}
$$</span></p>

<p>holds for <span class="math-container">$ 0 &lt; t &lt; 1$</span>. This is actually fairly easy to show</p>

<p><span class="math-container">\begin{eqnarray}
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &amp;=&amp; \gamma[t p_{t,1} + (1 - t)p_{t,2}]^2 + (t Y_1 + (1 - t)Y_{t,2})^2 \\
&amp;\vdots &amp; \\
&amp;=&amp; t [\gamma p_{t,1}^2 + Y_{t,1}^2] + (1 - t) [\gamma p_{t,2}^2 + Y_{t,2}^2] \\
&amp;&amp;\quad + t(1 - t) [\gamma(p_{t,1} - p_{t,2})^2 + (Y_{t,1} - Y_{t,2})^2] \\
&amp;=&amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\
&amp;&amp;\quad \underbrace{t(1 - t)L_t(p_{t,1} - p _{t_2}, Y_{t,1} - Y_{t,2})}_{&gt; 0} \\
&amp;&lt; &amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B3}
\end{eqnarray}</span></p>

<p>So <span class="math-container">$L_t$</span> is convex. To show that indifference curves are ellipses, just fix <span class="math-container">$L_t$</span> to a constant value, let's say <span class="math-container">$A &gt; 0$</span> then we have</p>

<p><span class="math-container">$$
A = \gamma p_t^2+ Y_t^2 ~~\Rightarrow~~ \frac{p_t^2}{A/\gamma} + \frac{Y_t}{A} = 1 \tag{B4}
$$</span></p>

<p>which are just ellipses centered at <span class="math-container">$p_t = 0 = \pi_t - \pi_t^\otimes$</span> and <span class="math-container">$Y_t = 0$</span> with axis lengths <span class="math-container">$\sqrt{A/\gamma}$</span> and <span class="math-container">$\sqrt{A}$</span></p>

CC BY-SA 4.0
2019-01-06T16:33:51.990Z
2019-01-07T00:20:51.153Z
2019-01-07T00:20:51.153Z
caverac
<p><a href="https://i.sstatic.net/ydABa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ydABa.png" alt="enter image description here" /></a></p>
<blockquote>
<p><span class="math-container">$$
L_t = \gamma(\pi_t - \pi_t^\otimes)^2 + \hat{Y}_t^2
$$</span></p>
</blockquote>
<p>Central banks loss function is given by the equation above. This loss is increasing and convex in the distance from the inflation target, i.e. the marginal loss increases at a faster rate than the increase in the distance from the target.</p>
<p>In Figure 1, we can visualise the preferences using indifference curves, where the ellipses have the center being the desired point (0, <span class="math-container">$\pi_t^\otimes$</span> = Target inflation point.) (saturation point - bliss point)</p>
<p>It is hard for me to see how the different ellipses are indifference curves, or convex from in the difference from target?. Any explanation would help a lot.</p>

<macroeconomics><monetary-policy><central-banking>
Central bank loss function
Niko
# Central bank loss function

(Asked by: username_0 on 2019-01-06)

[![enter image description here](https://i.sstatic.net/ydABa.png)](https://i.sstatic.net/ydABa.png)

> $$
> L\_t = \gamma(\pi\_t - \pi\_t^\otimes)^2 + \hat{Y}\_t^2
> $$

Central banks loss function is given by the equation above. This loss is increasing and convex in the distance from the inflation target, i.e. the marginal loss increases at a faster rate than the increase in the distance from the target.

In Figure 1, we can visualise the preferences using indifference curves, where the ellipses have the center being the desired point (0, $\pi\_t^\otimes$ = Target inflation point.) (saturation point - bliss point)

It is hard for me to see how the different ellipses are indifference curves, or convex from in the difference from target?. Any explanation would help a lot.

**Tags:** macroeconomics, monetary-policy, central-banking

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## 1 answers

### Answer by username_1 on 2019-01-06. Score: 3

> The intuitive way

Draw the function $L\_t$, the graph below is build with $\gamma = 0.8$ and $\pi\_t^\otimes = 1$

[![enter image description here](https://i.sstatic.net/MC0qI.png)](https://i.sstatic.net/MC0qI.png)

Convex in this context means that the function $L\_t$ looks like a bowl, just one well defined minimum. And if you cut it with a horizontal plane, the result is an ellipse.

[![enter image description here](https://i.sstatic.net/ISzGm.gif)](https://i.sstatic.net/ISzGm.gif)

> The formal way

Just for convenience, define the variable $p\_t = \pi\_t - \pi\_t^\otimes$ so the function becomes

$$
L\_t(p\_t, Y\_t) = \gamma p\_t^2 + Y\_t^2 \tag{B1}
$$

First note that if $\gamma > 0$ then $L\_t(p\_t, Y\_t) \ge 0$. To prove convexity you just need to show that for any pair $(p\_{t,1},Y\_{t,1})$ and $(p\_{t,2},Y\_{t,2})$ the condition

$$
L\_t(t p\_{t,1} + (1 - t)p\_{t,2}, t Y\_1 + (1 - t)Y\_{t,2}) < tL\_t(p\_{t,1}, Y\_t) + (1 -t)L\_t(p\_{t,2}, Y\_{t,2}) \tag{B2}
$$

holds for $ 0 < t < 1$. This is actually fairly easy to show

\begin{eqnarray}
L\_t(t p\_{t,1} + (1 - t)p\_{t,2}, t Y\_1 + (1 - t)Y\_{t,2}) &=& \gamma[t p\_{t,1} + (1 - t)p\_{t,2}]^2 + (t Y\_1 + (1 - t)Y\_{t,2})^2 \\
&\vdots & \\
&=& t [\gamma p\_{t,1}^2 + Y\_{t,1}^2] + (1 - t) [\gamma p\_{t,2}^2 + Y\_{t,2}^2] \\
&&\quad + t(1 - t) [\gamma(p\_{t,1} - p\_{t,2})^2 + (Y\_{t,1} - Y\_{t,2})^2] \\
&=& tL\_t(p\_{t,1}, Y\_t) + (1 -t)L\_t(p\_{t,2}, Y\_{t,2}) \\
&&\quad \underbrace{t(1 - t)L\_t(p\_{t,1} - p \_{t\_2}, Y\_{t,1} - Y\_{t,2})}\_{> 0} \\
&< & tL\_t(p\_{t,1}, Y\_t) + (1 -t)L\_t(p\_{t,2}, Y\_{t,2}) \tag{B3}
\end{eqnarray}

So $L\_t$ is convex. To show that indifference curves are ellipses, just fix $L\_t$ to a constant value, let's say $A > 0$ then we have

$$
A = \gamma p\_t^2+ Y\_t^2 ~~\Rightarrow~~ \frac{p\_t^2}{A/\gamma} + \frac{Y\_t}{A} = 1 \tag{B4}
$$

which are just ellipses centered at $p\_t = 0 = \pi\_t - \pi\_t^\otimes$ and $Y\_t = 0$ with axis lengths $\sqrt{A/\gamma}$ and $\sqrt{A}$


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