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How do you interpret this central bank’s loss function:

$$L(\hat x, \pi)=\alpha \hat{x}^2+(\pi-\pi^T)^2$$

where $\alpha >0$ and $\hat x$ is output gap.

Phillips curve is $\pi = \theta \hat{x}+\pi^T $

Sorry I am not economist, I am studying math, and I need its interpretation.

I guess I need to interpret the value of $a$, and I need to look at the geometry of the loss function in the Phillips curve diagram.

What is its interpretation in economic view?

Thank you so much!

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2 Answers 2

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The loss function reflects the central bank’s policy objective (i.e. their preferences for stabilizing inflation and some real variable around target levels, in this case output), and it says that the central bank is concerned about inflation fluctuating from their target level ($\pi^{T}$) and their perception of the equilibrium output (here reflected in the output gap).

Now, because the objective of the central bank includes two factors (output and inflation), we can know from the loss function how concerned the central bank is about stabilizing the output gap from the size of coefficient $\alpha$. Indeed, at varying levels of $\alpha$ it is easy to see that the reaction of the central bank to the same output gap will differ. Since $\alpha$ > 0 we know that they are concerned about the welfare loss associated with the output gap (if = 0 they would simply not be concerned about this, and if < 0 this would reflect that bigger output gaps are best since the function is to be minimized).

I can imagine that when looking at the geometry of the loss function in the Phillips curve diagram this may become clearer to you as someone from maths.

And just to note that once you derive the policy rule - the slope of the MR curve - (as in your other question), you will see that this policy rule not only depends on the central bank’s preferences $\alpha$ but also on the behavior of the private sector, the price flexibility as given by the parameter $\theta$.

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  • $\begingroup$ In addition to what you said, I also make some interpretation by using the Phillips curve diagram. Are the following statements correct? $\endgroup$
    – studentp
    Oct 20, 2019 at 10:34
  • $\begingroup$ “ $\alpha =1$ means that each indifference curve is a circle with $(x_e, \pi^T)$ at its centre. The loss declines as the circle gets smaller. $\endgroup$
    – studentp
    Oct 20, 2019 at 10:35
  • $\begingroup$ When $\pi =\pi^T$ and $x_e=x$, the circle shrinks to a single point (bliss point) and the loss is at minimum at zero. With $\alpha =1$, the central bank is indifferent between inflation 1% above (or below) $\pi^T$ and output 1% below (or above) $x_e$. They are on the same loss circle. $\endgroup$
    – studentp
    Oct 20, 2019 at 10:35
  • $\begingroup$ Only when $\alpha =1$ we have indifference circles. But when $\alpha <1$ then the central bank the central bank is indifferent between (say) inflation1% above (or below) $\pi^T$ and output 2% above (or below) $x_e$. This makes the indifference curves ellipsoid. And finally $\alpha >1$ means that central bank with less aversion to inflation will have ellipsoid indifference curves with a vertical rather than a horizontal orientation. $\endgroup$
    – studentp
    Oct 20, 2019 at 10:35
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    $\begingroup$ Yes, this holds perfectly for your example! I read it wrong and was about to comment something but then realised that you saw it perfectly since when 𝛼<1 this means that the central bank is inflation averse and when $\alpha > 1$, you can see that the central bank is unemployment-averse. Moreover, as you noted, utility declines with distance from the bliss-point. $\endgroup$
    – Ali
    Oct 20, 2019 at 12:19
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In this case, you can interpret this as the central banks' having two mandates: output gap and inflation target.

Here, the best possible scenario that would minimize the CB's loss function is having zero output gap ($\hat x^2 = 0$) and inflation is at the target ($\pi = \pi^T$). The CB simply wants to get as close to this as it could, with the "importance" of the output gap component given by $\alpha$. (What would $\alpha < 1$ mean, for example?)

I'm not sure how to help with the geometry part, but note that you can plug $\pi(\hat x)$ from the Phillips curve into the loss function.

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  • $\begingroup$ No no I think this is not answer that I need. Because what you say is related to another question that I linked. For example what is mean $\alpha$ without any calculation ? $\endgroup$
    – studentp
    Oct 20, 2019 at 7:53

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