<p>Just thinking out loud here, not saying this is correct others may have a better idea. I'm going to roll with this assumption that you are using best econometric practices so that you trust your estimates and want to take action based on them. </p>

<p>(1) The partial effect of going from $I=0$ to $I=1$ would be
$E(Y_1|I=1,N=\overline{N})-E(Y_1|I=0,N=\overline{N})$</p>

<p>$=\hat{\alpha_3}+\hat{\alpha_4}+\hat{\alpha_5}\overline{N}$</p>

<p>When irrigation is already positive, the marginal effect of irrigation on yields just be </p>

<p>$\frac{\partial E(Y_1|N,I)}{\partial I}=\hat{\alpha_3}+2\hat{\alpha_4}I+\hat{\alpha_5}N$</p>

<p>(2) So you are curious what level of Nitrogen makes it so that the marginal effect of nitrogen is the same as the partial effect of irrigation? I think you could get at this by finding the $I^*$ and $N^*$ such that</p>

<p>$\frac{\partial E(Y_1|N,I)}{\partial I}=\frac{\partial E(Y_1|N,I)}{\partial N}$?</p>

<p>We've already found $\frac{\partial E(Y_1|N,I)}{\partial I}$ above, so we need to calculate
 $\frac{\partial E(Y_1|N,I)}{\partial N}$ </p>

<p>$=\hat{\alpha_1}+2\hat{\alpha_2}N+\hat{\alpha_5}I$</p>

<p>From here $\frac{\partial E(Y_1|N,I)}{\partial I}|_{N=N^*,I=I^*}=\frac{\partial E(Y_1|N,I)}{\partial N}|_{N=N^*, I=I^*}$ </p>

<p>$\hat{\alpha_3}+2\hat{\alpha_4}I^*+\hat{\alpha_5}N^*=\hat{\alpha_1}+2\hat{\alpha_2}N^*+\hat{\alpha_5}I^*$</p>

<p>$\rightarrow (\hat{\alpha_5}-2\hat{\alpha_2})N^*=(\hat{\alpha_1}-\hat{\alpha_3})+(\hat{\alpha_5}-+2\hat{\alpha_4})I^*$</p>

<p>$\rightarrow N^*=\frac{(\hat{\alpha_1}-\hat{\alpha_3})+(\hat{\alpha_5}-+2\hat{\alpha_4})I^*}{(\hat{\alpha_5}-2\hat{\alpha_2})}$</p>

<p>I bet the optimal irrigation and fertilizer decision would equate the marginal product of fertilizer per dollar spent on fertilizer with the marginal product of irrigation per dollar spent on irrigation which would be</p>

<p>$\frac{\hat{\alpha_3}+\hat{\alpha_4}+\hat{\alpha_5}N}{P_{irrigation}}=\frac{\hat{\alpha_1}+2\hat{\alpha_2}N+\hat{\alpha_5}I}{P_{fertilizer}}$</p>

CC BY-SA 3.0
2016-12-01T16:06:03.013Z
2016-12-01T16:06:03.013Z
Hessian
<p>I will try to explain my question using two production functions here.</p>

<p>Let</p>

<p>$Y$ = Yield of a certain crop (tons/hectare)</p>

<p>Assume yield (output) is a function of two inputs, $Y = f(N,I)$, where</p>

<p>$N$ = Nitrogen fertilizer (input) (kg/hectare)</p>

<p>$I$ = Irrigation (mm/hectare)</p>

<p>Case 1: Assume I have data on 400 different plots. Each observation includes yield (output) and N (input). There is no irrigation in this case, $I = 0$.</p>

<p>$Y_1$ = $\alpha_0$ + $\alpha_1$ * $N_1$ + $\alpha_2$*$N^2$ + $e$  </p>

<p>Case 2: Irrigation (say up to 100 mm/hectare) is introduced in all plots, therefore, the value of $I$ ranges from 0-100. Irrigation increases yields in most plots. However, for a large number of plots, irrigation alone does not increase yields and additional N is required to increase yields. I have data on both inputs and the corresponding yields (output). </p>

<p>$Y_1$ = $\alpha_0$ + $\alpha_1$ * $N_1$ + $\alpha_2$*$N^2$ + $\alpha_3$ * $I_1$ + $\alpha_4$ * $I^2$+ $\alpha_5$ * $N*I$ + $e$ </p>

<p>What I want to measure is:
(1) What is the impact of irrigation on yields?
(2) Given that irrigation increases yields, how much additional Nitrogen (on average) is necessary to get to these yield increases? </p>

<p>a) From what I understand, I can't use a diff-in-diff type estimation here because the impacts of irrigation are heterogeneous, i.e., I can't simply add an irrigation dummy to the regression.</p>

<p>I would appreciate any help. Thank you so much! </p>

<regression><agriculture>
How do I calculate the impact of an independent variable on a dependent var. when the independent var. changes from 0 to some positive value?
Jj S
# How do I calculate the impact of an independent variable on a dependent var. when the independent var. changes from 0 to some positive value?

(Asked by: username_0 on 2016-11-30)

I will try to explain my question using two production functions here.

Let

$Y$ = Yield of a certain crop (tons/hectare)

Assume yield (output) is a function of two inputs, $Y = f(N,I)$, where

$N$ = Nitrogen fertilizer (input) (kg/hectare)

$I$ = Irrigation (mm/hectare)

Case 1: Assume I have data on 400 different plots. Each observation includes yield (output) and N (input). There is no irrigation in this case, $I = 0$.

$Y\_1$ = $\alpha\_0$ + $\alpha\_1$ \* $N\_1$ + $\alpha\_2$\*$N^2$ + $e$

Case 2: Irrigation (say up to 100 mm/hectare) is introduced in all plots, therefore, the value of $I$ ranges from 0-100. Irrigation increases yields in most plots. However, for a large number of plots, irrigation alone does not increase yields and additional N is required to increase yields. I have data on both inputs and the corresponding yields (output).

$Y\_1$ = $\alpha\_0$ + $\alpha\_1$ \* $N\_1$ + $\alpha\_2$\*$N^2$ + $\alpha\_3$ \* $I\_1$ + $\alpha\_4$ \* $I^2$+ $\alpha\_5$ \* $N\*I$ + $e$

What I want to measure is:
(1) What is the impact of irrigation on yields?
(2) Given that irrigation increases yields, how much additional Nitrogen (on average) is necessary to get to these yield increases?

a) From what I understand, I can't use a diff-in-diff type estimation here because the impacts of irrigation are heterogeneous, i.e., I can't simply add an irrigation dummy to the regression.

I would appreciate any help. Thank you so much!

**Tags:** regression, agriculture

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## 1 answers

### Answer by username_1 on 2016-12-01. Score: 1

Just thinking out loud here, not saying this is correct others may have a better idea. I'm going to roll with this assumption that you are using best econometric practices so that you trust your estimates and want to take action based on them.

(1) The partial effect of going from $I=0$ to $I=1$ would be
$E(Y\_1|I=1,N=\overline{N})-E(Y\_1|I=0,N=\overline{N})$

$=\hat{\alpha\_3}+\hat{\alpha\_4}+\hat{\alpha\_5}\overline{N}$

When irrigation is already positive, the marginal effect of irrigation on yields just be

$\frac{\partial E(Y\_1|N,I)}{\partial I}=\hat{\alpha\_3}+2\hat{\alpha\_4}I+\hat{\alpha\_5}N$

(2) So you are curious what level of Nitrogen makes it so that the marginal effect of nitrogen is the same as the partial effect of irrigation? I think you could get at this by finding the $I^\*$ and $N^\*$ such that

$\frac{\partial E(Y\_1|N,I)}{\partial I}=\frac{\partial E(Y\_1|N,I)}{\partial N}$?

We've already found $\frac{\partial E(Y\_1|N,I)}{\partial I}$ above, so we need to calculate
$\frac{\partial E(Y\_1|N,I)}{\partial N}$

$=\hat{\alpha\_1}+2\hat{\alpha\_2}N+\hat{\alpha\_5}I$

From here $\frac{\partial E(Y\_1|N,I)}{\partial I}|\_{N=N^\*,I=I^\*}=\frac{\partial E(Y\_1|N,I)}{\partial N}|\_{N=N^\*, I=I^\*}$

$\hat{\alpha\_3}+2\hat{\alpha\_4}I^\*+\hat{\alpha\_5}N^\*=\hat{\alpha\_1}+2\hat{\alpha\_2}N^\*+\hat{\alpha\_5}I^\*$

$\rightarrow (\hat{\alpha\_5}-2\hat{\alpha\_2})N^\*=(\hat{\alpha\_1}-\hat{\alpha\_3})+(\hat{\alpha\_5}-+2\hat{\alpha\_4})I^\*$

$\rightarrow N^\*=\frac{(\hat{\alpha\_1}-\hat{\alpha\_3})+(\hat{\alpha\_5}-+2\hat{\alpha\_4})I^\*}{(\hat{\alpha\_5}-2\hat{\alpha\_2})}$

I bet the optimal irrigation and fertilizer decision would equate the marginal product of fertilizer per dollar spent on fertilizer with the marginal product of irrigation per dollar spent on irrigation which would be

$\frac{\hat{\alpha\_3}+\hat{\alpha\_4}+\hat{\alpha\_5}N}{P\_{irrigation}}=\frac{\hat{\alpha\_1}+2\hat{\alpha\_2}N+\hat{\alpha\_5}I}{P\_{fertilizer}}$


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