<p>My understanding is, IRF of output (<span class="math-container">$Y$</span>) to monetary shock (<span class="math-container">$\epsilon$</span>) is a set of partial derivatives, i.e., IRF of output (<span class="math-container">$Y$</span>) to monetary shock (<span class="math-container">$\epsilon$</span>) is {<span class="math-container">$\partial Y_{t+j}/ \partial \epsilon_t, j = 0,1,2,3,4...$</span>}.</p>
<p>If expansionary monetary shock occurs at time <span class="math-container">$t$</span>, then, change in monetary shock at time <span class="math-container">$t$</span> is negative <span class="math-container">$\partial \epsilon_t&lt;0$</span>. Then, if IRF of output at time <span class="math-container">$t$</span> is negative, shouldn't we conclude &quot;output increases at time <span class="math-container">$t$</span> with decrease in the interest rate&quot; instead of &quot;output decreases at time <span class="math-container">$t$</span> with decrease in the interest rate.&quot; I am asking this question because all the papers I read say, &quot;output decreases at time <span class="math-container">$t$</span> with decrease in the interest rate.&quot;</p>

CC BY-SA 4.0
2024-10-05T06:17:05.713Z
Your reading of the literature is mistaken. The standard literature (e.g. a host of papers by Larry Christiano) on the effect of monetary policy shows that output falls if interest rates increase.
jpfeifer
CC BY-SA 4.0
2024-10-06T00:26:20.340Z
Thank you so much @jpfeifer I wrote this oppositely. Silly me. Would it possible to help me out with a question.  Should the IRF of consumption to negative productivity shock be negative or positive? I would think negative but I am little bit confused because IRF is a set of partial derivatives correct? Then, if ∂et <0 and since shock decreases consumption, ∂Ct<0 , so, ∂Ct/∂et have to be > 0. What am doing I wrong? $C_t$ is consumption and $e_t$ is productivity shock.
Hyun Jung
CC BY-SA 4.0
2024-10-07T08:27:47.263Z
The fraction $\partial c/\partial \varepsilon$ only give the sign. You still need to multiply it with the actual shock you are considering. The total effect is $\partial c/\partial \varepsilon\times \Delta \varepsilon$. The partial derivative is usually positive (C increases after a positive shock), but given that you multiply the expression by a negative shock you are considering, the overall effect/IRF is negative after a negative shock.
jpfeifer
<macroeconomics><econometrics><monetary-policy>
Question regarding impulse response functions (IRFs)
Hyun Jung
# Question regarding impulse response functions (IRFs)

(Asked by: username_0 on 2024-10-05)

My understanding is, IRF of output ($Y$) to monetary shock ($\epsilon$) is a set of partial derivatives, i.e., IRF of output ($Y$) to monetary shock ($\epsilon$) is {$\partial Y\_{t+j}/ \partial \epsilon\_t, j = 0,1,2,3,4...$}.

If expansionary monetary shock occurs at time $t$, then, change in monetary shock at time $t$ is negative $\partial \epsilon\_t<0$. Then, if IRF of output at time $t$ is negative, shouldn't we conclude "output increases at time $t$ with decrease in the interest rate" instead of "output decreases at time $t$ with decrease in the interest rate." I am asking this question because all the papers I read say, "output decreases at time $t$ with decrease in the interest rate."

**Tags:** macroeconomics, econometrics, monetary-policy

> **Comments:**
> **username_1**: Your reading of the literature is mistaken. The standard literature (e.g. a host of papers by Larry Christiano) on the effect of monetary policy shows that output falls if interest rates increase.  
> **username_0**: Thank you so much @username_1 I wrote this oppositely. Silly me. Would it possible to help me out with a question.  Should the IRF of consumption to negative productivity shock be negative or positive? I would think negative but I am little bit confused because IRF is a set of partial derivatives correct? Then, if ∂et <0 and since shock decreases consumption, ∂Ct<0 , so, ∂Ct/∂et have to be > 0. What am doing I wrong? $C_t$ is consumption and $e_t$ is productivity shock.  
> **username_1**: The fraction $\partial c/\partial \varepsilon$ only give the sign. You still need to multiply it with the actual shock you are considering. The total effect is $\partial c/\partial \varepsilon\times \Delta \varepsilon$. The partial derivative is usually positive (C increases after a positive shock), but given that you multiply the expression by a negative shock you are considering, the overall effect/IRF is negative after a negative shock.  

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## 0 answers

