<p>Yes they are the same definitions. Graphically the different integrals calculate the same area just rotated 90 degrees. Geometric shapes have the same area regardless of rotation.</p>
<p>General Proof:</p>
<p>Suppose we have a region of <strong>fixed</strong> size called CS in I quadrant of Cartesian coordinate system enclosed by <span class="math-container">$p(q)$</span> and <span class="math-container">$p= p_0$</span>. The area of the region would be given by:</p>
<p><span class="math-container">$$\text{ CS} = \lim_{n \to \infty} \sum [p(q_i)- p_0] \Delta q, \text{with } \quad \Delta q = \frac{q_0-0}{n}  $$</span></p>
<p>Taking the limit we get:</p>
<p><span class="math-container">$$ \text{CS} =   \int_{0}^{q_0} [p(q) - p_0] \ dq$$</span></p>
<p>Now integrate the same <strong>fixed</strong> area CS bounded by <span class="math-container">$p(q)$</span> and <span class="math-container">$p_0$</span> by <span class="math-container">$y$</span> axis.</p>
<p>Since CS is bounded by <span class="math-container">$p(q)$</span> and <span class="math-container">$p_0$</span> on <span class="math-container">$y$</span> axis we are integrating over interval <span class="math-container">$[p_{max},p_0]$</span> over inverse function <span class="math-container">$p(q)^{-1}$</span> i.e. <span class="math-container">$q=p(q)$</span>  (we do not include <span class="math-container">$p_0$</span> as it is constant function and thus the boundary on <span class="math-container">$y$</span> axis). Hence, on <span class="math-container">$y$</span> axes the <strong>fixed</strong> area of <span class="math-container">$CS$</span> is defined as:</p>
<p><span class="math-container">$$\text{CS}= \lim_{n \to \infty} \sum q(p_i)\Delta p, \text{with} \quad \Delta p = \frac{p_{max}-p_0}{n}$$</span></p>
<p>taking the limit we get:</p>
<p><span class="math-container">$$\text{CS}= \int_{p_0}^{p_{max}}q(p)dp$$</span></p>
<p>Since we are talking about <span class="math-container">$CS$</span> of the same size we have:</p>
<p><span class="math-container">$$ \int_{0}^{q_0} [p(q) - p_0] \ dq = \text{CS} = \int_{p_0}^{p_{max}}q(p)dp$$</span></p>
<p>And hence we proven that:</p>
<p><span class="math-container">$$ \int_{0}^{q_0} [p(q) - p_0] \ dq  = \int_{p_0}^{p_{max}}q(p)dp$$</span></p>
<hr />
<p>Example:</p>
<p>You can see that by trying various demand functions. Suppose we have demand given by <span class="math-container">$Q=100-p$</span>, equilibrium price is given by <span class="math-container">$p=10$</span>.</p>
<p>Varian:</p>
<p><span class="math-container">$$CS(10)= \int_{10}^{100}(100-p)dp= 100(100) -\frac{1}{2}(100)^2-100(10) -\frac{1}{2}(10)^2=4050$$</span></p>
<p>Non-Varian:</p>
<p>we first have to solve for inverse demand which is given by: <span class="math-container">$p=100-q$</span> (recall that fixing $p=10 \implies q^*=90).</p>
<p><span class="math-container">$$CS(10)= \int_{0}^{90}[(100-q)-10]dp= 4050$$</span></p>
<p>This will hold for any arbitrary function since consumer surplus is graphically a geometrical object and the respective integrals are just different ways of calculating area of this object depending on how you rotate the plane with the object.</p>
<p>More generally what Varian approach is doing would be known as integrating along the y-axis.</p>

CC BY-SA 4.0
2022-10-27T17:33:05.147Z
Can you please help me show that mathematically? I understand the rotation, but I don't see how to write that in a more rigorous manner.
marblegoolies
CC BY-SA 4.0
2022-10-27T18:52:07.340Z
@marblegoolies I updated the question with general proof for arbitrary p(q) and p_0, (of course assuming p(q) is continuous, inverse exists and is also continuous)
1muflon1
CC BY-SA 4.0
2022-10-27T21:00:09.320Z
Is the statement also true if $q(p_{max} ) > 0$? If not, do you mind pointing out where in your proof you use this assumption?
Giskard
CC BY-SA 4.0
2022-10-27T21:23:23.683Z
@Giskard but by common definition $p_{max}$ is $p$ that solves $q(p)=0$. This is how any undergraduate text uses p_max so how could $q(p_{max})>0$? That’s simply implicitly not allowed the same as q can’t be less than 0
1muflon1
CC BY-SA 4.0
2022-10-27T21:57:10.643Z
1. Not sure this is the definition of $p_{max}$ OP has in mind.
Giskard
CC BY-SA 4.0
2022-10-27T21:57:39.550Z
2. This is a rigorous mathematical proof, right? Are you using the assumption of $q(p_{max}) = 0$ or not? If yes, where?
Giskard
CC BY-SA 4.0
2022-10-27T22:02:21.573Z
@Giskard 1. huh? Literally quoting from op " q(pmax)=0." 2. Yes I am using it. Its already mentioned in Q and its extremely common assumption so I do not repeat it explicitly again in A
1muflon1
CC BY-SA 4.0
2022-10-27T22:03:00.080Z
Okay, I thought about 1. and I concede the point.
Giskard
CC BY-SA 4.0
2022-10-27T22:05:19.573Z
And I also concede 2. I have read this question on my phone, and misunderstood it quite a bit!
Giskard
CC BY-SA 4.0
2022-10-27T22:06:10.203Z
@Giskard no problem, always happy to have you peer review my posts so I can improve them further, mistakes sometimes happen nobody is perfect
1muflon1
CC BY-SA 4.0
2022-10-27T22:06:22.740Z
Would prefer your answer if the general proof came before the example, but to each her own. There you go, peer review (:
Giskard
CC BY-SA 4.0
2022-10-27T22:08:20.723Z
@Giskard good point, I switched it around, I first just had an example since I thought OP is just not sure it works and I only added general proof as an edit but makes sense to put it first
1muflon1
CC BY-SA 4.0
2022-10-27T22:09:49.883Z
@Giskard PS: off topic, but you might be happy about this https://economics.meta.stackexchange.com/questions/2198/should-we-reduce-the-number-of-votes-required-to-close-from-5-to-3-close-votes
1muflon1
CC BY-SA 4.0
2022-10-28T06:32:38.067Z
Very much appreciated!
marblegoolies
CC BY-SA 4.0
2022-10-27T15:58:42.283Z
2022-10-27T22:07:44.477Z
2022-10-27T22:07:44.477Z
1muflon1
<p>Varian defines consumer surplus at price <span class="math-container">$p = p_0$</span> as <span class="math-container">$\displaystyle CS(p_0) = \int_{p_0}^{p_\max} q(p) \ dp$</span> where <span class="math-container">$p_\max$</span> exists and <span class="math-container">$q(p_\max) = 0$</span>.</p>
<p>The other definition that I generally use is <span class="math-container">$\displaystyle CS(p_0) = \int_{0}^{q_0} [p(q) - p_0] \ dq$</span>.</p>
<p>Are the two definitions same? I tried equating them and I couldn't prove that they are equal. Is the second definition incorrect?</p>

<microeconomics><consumer-surplus>
Consumer surplus definition
marblegoolies
# Consumer surplus definition

(Asked by: username_0 on 2022-10-27)

Varian defines consumer surplus at price $p = p\_0$ as $\displaystyle CS(p\_0) = \int\_{p\_0}^{p\_\max} q(p) \ dp$ where $p\_\max$ exists and $q(p\_\max) = 0$.

The other definition that I generally use is $\displaystyle CS(p\_0) = \int\_{0}^{q\_0} [p(q) - p\_0] \ dq$.

Are the two definitions same? I tried equating them and I couldn't prove that they are equal. Is the second definition incorrect?

**Tags:** microeconomics, consumer-surplus


---

## 1 answers

### Answer by username_1 on 2022-10-27. Score: 1 | Accepted answer

Yes they are the same definitions. Graphically the different integrals calculate the same area just rotated 90 degrees. Geometric shapes have the same area regardless of rotation.

General Proof:

Suppose we have a region of **fixed** size called CS in I quadrant of Cartesian coordinate system enclosed by $p(q)$ and $p= p\_0$. The area of the region would be given by:

$$\text{ CS} = \lim\_{n \to \infty} \sum [p(q\_i)- p\_0] \Delta q, \text{with } \quad \Delta q = \frac{q\_0-0}{n} $$

Taking the limit we get:

$$ \text{CS} = \int\_{0}^{q\_0} [p(q) - p\_0] \ dq$$

Now integrate the same **fixed** area CS bounded by $p(q)$ and $p\_0$ by $y$ axis.

Since CS is bounded by $p(q)$ and $p\_0$ on $y$ axis we are integrating over interval $[p\_{max},p\_0]$ over inverse function $p(q)^{-1}$ i.e. $q=p(q)$ (we do not include $p\_0$ as it is constant function and thus the boundary on $y$ axis). Hence, on $y$ axes the **fixed** area of $CS$ is defined as:

$$\text{CS}= \lim\_{n \to \infty} \sum q(p\_i)\Delta p, \text{with} \quad \Delta p = \frac{p\_{max}-p\_0}{n}$$

taking the limit we get:

$$\text{CS}= \int\_{p\_0}^{p\_{max}}q(p)dp$$

Since we are talking about $CS$ of the same size we have:

$$ \int\_{0}^{q\_0} [p(q) - p\_0] \ dq = \text{CS} = \int\_{p\_0}^{p\_{max}}q(p)dp$$

And hence we proven that:

$$ \int\_{0}^{q\_0} [p(q) - p\_0] \ dq = \int\_{p\_0}^{p\_{max}}q(p)dp$$

---

Example:

You can see that by trying various demand functions. Suppose we have demand given by $Q=100-p$, equilibrium price is given by $p=10$.

Varian:

$$CS(10)= \int\_{10}^{100}(100-p)dp= 100(100) -\frac{1}{2}(100)^2-100(10) -\frac{1}{2}(10)^2=4050$$

Non-Varian:

we first have to solve for inverse demand which is given by: $p=100-q$ (recall that fixing $p=10 \implies q^\*=90).

$$CS(10)= \int\_{0}^{90}[(100-q)-10]dp= 4050$$

This will hold for any arbitrary function since consumer surplus is graphically a geometrical object and the respective integrals are just different ways of calculating area of this object depending on how you rotate the plane with the object.

More generally what Varian approach is doing would be known as integrating along the y-axis.

> **Comments:**
> **username_0**: Can you please help me show that mathematically? I understand the rotation, but I don't see how to write that in a more rigorous manner.  
> **username_1**: @username_0 I updated the question with general proof for arbitrary p(q) and p_0, (of course assuming p(q) is continuous, inverse exists and is also continuous)  
> **username_2**: Is the statement also true if $q(p_{max} ) > 0$? If not, do you mind pointing out where in your proof you use this assumption?  
> **username_1**: @username_2 but by common definition $p_{max}$ is $p$ that solves $q(p)=0$. This is how any undergraduate text uses p_max so how could $q(p_{max})>0$? That’s simply implicitly not allowed the same as q can’t be less than 0  
> **username_2**: 1. Not sure this is the definition of $p_{max}$ OP has in mind.  
> **username_2**: 2. This is a rigorous mathematical proof, right? Are you using the assumption of $q(p_{max}) = 0$ or not? If yes, where?  
> **username_1**: @username_2 1. huh? Literally quoting from op " q(pmax)=0." 2. Yes I am using it. Its already mentioned in Q and its extremely common assumption so I do not repeat it explicitly again in A  
> **username_2**: Okay, I thought about 1. and I concede the point.  
> **username_2**: And I also concede 2. I have read this question on my phone, and misunderstood it quite a bit!  
> **username_1**: @username_2 no problem, always happy to have you peer review my posts so I can improve them further, mistakes sometimes happen nobody is perfect  
> **username_2**: Would prefer your answer if the general proof came before the example, but to each her own. There you go, peer review (:  
> **username_1**: @username_2 good point, I switched it around, I first just had an example since I thought OP is just not sure it works and I only added general proof as an edit but makes sense to put it first  
> **username_1**: @username_2 PS: off topic, but you might be happy about this https://economics.meta.stackexchange.com/questions/2198/should-we-reduce-the-number-of-votes-required-to-close-from-5-to-3-close-votes  
> **username_0**: Very much appreciated!  

---

