# How to Calculate the productivity multiplier?

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 $$\frac{1}{1-\alpha}$$.

• You should show any work you've already done in attempting to solve this question. – Kenny LJ Jan 27 '19 at 2:30
• @KennyLJ I updated with my attempt – Fatima Jan 27 '19 at 2:54

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 $$A$$: $$\frac{\partial y^*}{\partial A}\div \frac{y^*}{A}=\frac{1}{1-\alpha}.$$