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I have to find the GDP in equilibrium for a IS LM model. It is given that

$M^d (Y,r)=M_0+M_1Y-M_2r$ and $M^d=M/P$, $M_0,M_1,M_2>0$ and $M^d$ is money demand.

my solution so far

I have found that $IS=\frac{1}{1-b}(a-bT+I_0-I_1r+G)$ and $a,b,c,I_0,I_1>0$, $0<b<1$ from an previous result. Then i proceeded to find $LM=-\frac{M_0}{M_1}+\frac{M}{M_1P}+\frac{M_2r}{M_1}$ (we have to have $Y$ as a function of $r$ and not the usual reversed way). $G,P,M,P$ are all exo.

Now, I do not know how the GDP is derived from equilibrium when we have to express in terms of $r$.

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Assuming there were no mistakes in IS and LM calculation you just solve one for $r$ substitute into another and solve for output - this is no different then solving a system of two equations.

$$Y = -\frac{M_0}{M_1}+\frac{M}{M_1P}+\frac{M_2r}{M_1} \implies r = \frac{M_1}{M_2}\left( Y +\frac{M_0}{M_1}-\frac{M}{M_1P}\right)$$

just plug this expression for $r$ into the IS and you get the solution.

$$Y = \frac{1}{1-b}(a-bT+I_0-I_1 \frac{M_1}{M_2}\left( Y +\frac{M_0}{M_1}-\frac{M}{M_1P}\right)+G) $$

$$Y^* = \frac{(1-b)M_2}{((1-b)M_2+ I_1)M_1} \frac{1}{1-b}(a-bT+I_0 -I_1 \frac{M_1}{M_2}\left(\frac{M_0}{M_1}-\frac{M}{M_1P}\right)+G)\\ =\frac{M_2}{((1-b)M_2+ I_1)M_1} (a-bT+I_0 - \frac{I_1}{M_2}\left( M_0-\frac{M}{P}\right)+G)$$

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  • $\begingroup$ Yes this was my idea as well. I must say the equation looks kind of strange and I will have to double check my calculations for IS and LM. For LM I just had that $M_0+M_1Y-M_2r=M/P$ which is just subtracting everything else than $M_1Y$ and then dividing through by $M_1$. For IS I have that $C=a+b(Y-T) \;\; \wedge \;\; I(r)=I_0-I_1 r$ and that $Y=C+I+G$. Substituting thoes into $Y$ yields (I think) the above written IS. $\endgroup$ – user31331 Dec 26 '20 at 15:00
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    $\begingroup$ @bymathformath you can double check the IS and LM but it does not seem anything out of ordinary - see for example, Blanchard et al Macroeconomics where they also get large expression - here expression is complicated mainly by having all those multiple $M$ aggregates instead of just having one $M$. In any case in the textbook you will see that this is how you solve these sort of problems. Also, if you think the above solved your issue consider accepting it $\endgroup$ – 1muflon1 Dec 26 '20 at 15:03
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    $\begingroup$ @bymathformath in short, because in the full model multiplier also depends on the investment spending (this also happens in some goods market equilibrium models if you assume I(Y)) and in addition some of the effect (real) government spending would have would be offset by increase in money supply (and its further effects on price level etc) since in the model above when output increases people’s demand for money increases as well. $\endgroup$ – 1muflon1 Dec 26 '20 at 16:59
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    $\begingroup$ @bymathformath what do you mean? It can be less than 1 or more than one. If b=0.9,M2=10, I=1 and M1=1 multiplier will be 5 if b=0.9,M2=1,I=1, M1=1 multiplier will be approximately 0.91 so clearly it can be either higher than 1 or less than 1 depending on parameters $\endgroup$ – 1muflon1 Dec 27 '20 at 17:22
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    $\begingroup$ @bymathformath I don’t know details of your assignment there might be other assumptions about parameter values there $\endgroup$ – 1muflon1 Dec 27 '20 at 17:24

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