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Given a Cobb Douglas

$Y_t = A (K_t^\alpha L_t^{1-\alpha}) $

$ K_{t+1} = sY_t + (1-\delta) K_t$

How do we get the multiplier on productivity to be equal to $ \frac{1}{1-\alpha}$? I understand that if productivity increases, output increases, thus we get more capital and thereby more output and so on. But I can't reach this multiplier.

My Attempt:

If have x increase in A, then $Y_t=(1+x)A (K_t^\alpha L_t^{1-\alpha}) $

That is an x increase in Y.

$ K_{t+1} = s(1+x)A (K_t^\alpha L_t^{1-\alpha}) + (1-\delta) K_t$

$Y_{t+1}=(1+x)A * K_{t+1}^\alpha *L_{t+1}^{1-\alpha} $ $Y_{t+1}= (1+x)A * (s(1+x)A (K_t^\alpha L_t^{1-\alpha}) + (1-\delta) K_t)^\alpha * L_{t+1}^{1-\alpha} $

I think the increase here should be $x * \alpha $ but I can't see it. so that for a unit increase in productivity i.e x=1; $\Delta Y = 1 + \alpha + \alpha^2 +... =\frac{1}{1-\alpha} $

Also in the steady state we have, $ Y = A^\frac{1}{1-\alpha} * \frac{s}{\delta}^\frac{\alpha}{1-\alpha} * L$ taking logs we have that percentage change in A increases Y by $\frac{1}{1-\alpha}$.

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    $\begingroup$ You should show any work you've already done in attempting to solve this question. $\endgroup$ – Kenny LJ Jan 27 at 2:30
  • $\begingroup$ @KennyLJ I updated with my attempt $\endgroup$ – Fatima Jan 27 at 2:54
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Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^\alpha$.

Steady state is given by: $$k^*=sy^*+(1-\delta)k^*,$$ or $$k^*=sA(k^*)^\alpha+(1-\delta)k^*.$$

Doing the algebra: $$k^*=\left(\frac{sA}{\delta}\right)^{\frac{1}{1-\alpha}}.$$

And: $$y^*=A\left(\frac{sA}{\delta}\right)^{\frac{\alpha}{1-\alpha}}=A^{\frac{1}{1-\alpha}}\left(\frac{s}{\delta}\right)^{\frac{\alpha}{1-\alpha}}.$$

I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$\frac{\partial y^*}{\partial A}=\frac{1}{1-\alpha}A^{\frac{\alpha}{1-\alpha}}\left(\frac{s}{\delta}\right)^{\frac{\alpha}{1-\alpha}}.$$

However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$\frac{\partial y^*}{\partial A}\div \frac{y^*}{A}=\frac{1}{1-\alpha}.$$

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