<p>I have 2 related questions that i think are straightforward but am worried I have over-simplified!</p>

<p>(i) Suppose the demand curve is downward sloping and the supply curve is upwards 
sloping. Suppose also that I have estimated the price elasticity of demand <span class="math-container">$ \frac{\partial ln(D)}{\partial ln(p)}  $</span>, </p>

<p>Can I obtain the responsiveness <em>of</em> price <em>to</em> demand simply by taking the reciprocal?</p>

<p><span class="math-container">$ \dfrac{1}{ \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(p)}{\partial ln(D)}  $</span></p>

<p>By analogy to discrete changes, the rearrangement below seems to work</p>

<p><span class="math-container">$ \dfrac{1}{ \frac{\Delta D}{\Delta p} \cdot\frac{p}{D}}  = \frac{\Delta p}{\Delta D} \cdot \frac{D}{p} $</span></p>

<p>Or do I also need an estimate of the price elasticity of supply?</p>

<p>(ii) Relatedly, now suppose I also have the price elasticity of supply, and that I want to estimate the responsiveness of the supply curve to shifts in the demand curve (with the mechanism being that some positive shock to the demand curve raises prices, which incentivises firms to supply more output). Is the following valid?</p>

<p><span class="math-container">$ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{  \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(S)}{\partial ln(p)} \cdot \frac{\partial ln(p)}{\partial ln(D)} = \frac{\partial ln(S)}{\partial ln(D)}  $</span></p>

<p>Thanks loads for your help, and apologies if I missed a similar question</p>

CC BY-SA 4.0
2019-02-22T21:47:33.843Z
Perhaps the following exercise will help you order your thoughts: Suppose $D(p) = 100 - p$, $S(p) = 3p$. Now there is a shift in the demand curve, and it becomes $\hat{D}(p) = 120 - p$. (i) In this case, what do you denote by $\Delta D$, or $\frac{\partial ln(p)}{\partial ln(D)} $? (ii) In this case what do you denote by $\frac{\partial ln(S)}{\partial ln(D)} $?
Giskard
CC BY-SA 4.0
2019-02-22T22:39:10.040Z
In this case, I would have thought (i) $ \Delta D = 20 $ and $ \frac{\partial ln(D)}{\partial ln(p)} = -1 $, thus $ \frac{\partial ln(p)}{\partial ln(D)} = 1/-1 = -1 $. This seems right because if I rearrange the demand curve to get $ p(D) = 100 - D $ and differentiate, I also get $ \frac{\partial ln(p)}{\partial ln(D)} = -1. $ (ii) would be $ 3 * -1 = -3. $ . Is this right?
leafnoise
CC BY-SA 4.0
2019-02-22T22:47:23.087Z
No. Could you please detail your calculations? The numbers you gave seem pretty random. E.g. I see that $\frac{\partial D(p)}{\partial p} = -1$, but why do you think the same is true for $\frac{\partial ln(D)}{\partial ln(p)}$?
Giskard
CC BY-SA 4.0
2019-02-23T00:02:29.713Z
Right, yeah sorry i forgot to convert the differential into an elasticity. So (i) $  \frac{\partial D}{\partial p}  \cdot \frac{p}{D}  = -1 \cdot \frac{p}{D} = \frac{-p}{D} $. The reciprocal of this is $\frac{-D}{p}$. If i rearrange the demand curve to get $ p(D) = 100 - D$ and find the elasticity that way, i get $  \frac{\partial p}{\partial D}  \cdot \frac{D}{p}  = -1 \cdot \frac{D}{p} = \frac{-D}{p} $. Then, if my logic for (ii) is right, i would get $ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{\frac{\partial ln(D)}{\partial ln(p)}} = \dfrac{\frac{3p}{S}}{\frac{-p}{D}} = \frac{-3 D}{S}$
leafnoise
CC BY-SA 4.0
2019-02-23T08:32:01.457Z
"i rearrange the demand curve to get $p(D)=100−D$" This is pretty much the issue, you are denoting both the demand function (a function) and the quantity demanded at a certain price (a number) with $D$.
Giskard
CC BY-SA 4.0
2019-02-25T12:21:42.050Z
Not sure I fully understand that: $ p = p(D) = D(p)^{-1} $, i.e. just the inverse demand function, no?
leafnoise
CC BY-SA 4.0
2019-02-25T17:11:24.013Z
I am not sure what you do not understand. There is both a market price $p$ and an inverse demand function $p()$. These are not the same mathematical object, as one is a function and the other is not.
Giskard
<supply-and-demand>
Does the reciprocal of the price elasticity of demand have an interpretation?
leafnoise
# Does the reciprocal of the price elasticity of demand have an interpretation?

(Asked by: username_0 on 2019-02-22)

I have 2 related questions that i think are straightforward but am worried I have over-simplified!

(i) Suppose the demand curve is downward sloping and the supply curve is upwards
sloping. Suppose also that I have estimated the price elasticity of demand $ \frac{\partial ln(D)}{\partial ln(p)} $,

Can I obtain the responsiveness *of* price *to* demand simply by taking the reciprocal?

$ \dfrac{1}{ \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(p)}{\partial ln(D)} $

By analogy to discrete changes, the rearrangement below seems to work

$ \dfrac{1}{ \frac{\Delta D}{\Delta p} \cdot\frac{p}{D}} = \frac{\Delta p}{\Delta D} \cdot \frac{D}{p} $

Or do I also need an estimate of the price elasticity of supply?

(ii) Relatedly, now suppose I also have the price elasticity of supply, and that I want to estimate the responsiveness of the supply curve to shifts in the demand curve (with the mechanism being that some positive shock to the demand curve raises prices, which incentivises firms to supply more output). Is the following valid?

$ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{ \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(S)}{\partial ln(p)} \cdot \frac{\partial ln(p)}{\partial ln(D)} = \frac{\partial ln(S)}{\partial ln(D)} $

Thanks loads for your help, and apologies if I missed a similar question

**Tags:** supply-and-demand
> **Comments:**
> **username_1**: Perhaps the following exercise will help you order your thoughts: Suppose $D(p) = 100 - p$, $S(p) = 3p$. Now there is a shift in the demand curve, and it becomes $\hat{D}(p) = 120 - p$. (i) In this case, what do you denote by $\Delta D$, or $\frac{\partial ln(p)}{\partial ln(D)} $? (ii) In this case what do you denote by $\frac{\partial ln(S)}{\partial ln(D)} $?  
> **username_0**: In this case, I would have thought (i) $ \Delta D = 20 $ and $ \frac{\partial ln(D)}{\partial ln(p)} = -1 $, thus $ \frac{\partial ln(p)}{\partial ln(D)} = 1/-1 = -1 $. This seems right because if I rearrange the demand curve to get $ p(D) = 100 - D $ and differentiate, I also get $ \frac{\partial ln(p)}{\partial ln(D)} = -1. $ (ii) would be $ 3 * -1 = -3. $ . Is this right?  
> **username_1**: No. Could you please detail your calculations? The numbers you gave seem pretty random. E.g. I see that $\frac{\partial D(p)}{\partial p} = -1$, but why do you think the same is true for $\frac{\partial ln(D)}{\partial ln(p)}$?  
> **username_0**: Right, yeah sorry i forgot to convert the differential into an elasticity. So (i) $  \frac{\partial D}{\partial p}  \cdot \frac{p}{D}  = -1 \cdot \frac{p}{D} = \frac{-p}{D} $. The reciprocal of this is $\frac{-D}{p}$. If i rearrange the demand curve to get $ p(D) = 100 - D$ and find the elasticity that way, i get $  \frac{\partial p}{\partial D}  \cdot \frac{D}{p}  = -1 \cdot \frac{D}{p} = \frac{-D}{p} $. Then, if my logic for (ii) is right, i would get $ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{\frac{\partial ln(D)}{\partial ln(p)}} = \dfrac{\frac{3p}{S}}{\frac{-p}{D}} = \frac{-3 D}{S}$  
> **username_1**: "i rearrange the demand curve to get $p(D)=100−D$" This is pretty much the issue, you are denoting both the demand function (a function) and the quantity demanded at a certain price (a number) with $D$.  
> **username_0**: Not sure I fully understand that: $ p = p(D) = D(p)^{-1} $, i.e. just the inverse demand function, no?  
> **username_1**: I am not sure what you do not understand. There is both a market price $p$ and an inverse demand function $p()$. These are not the same mathematical object, as one is a function and the other is not.  

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