<blockquote>
<p>If the income increases by $d, will it mean the utility at the optimal point increases by λd?</p>
</blockquote>
<p>Yes it will. For a problem:</p>
<p><span class="math-container">$$\max U(x,y) \text{ s.t. } p_xx+ p_yy=I$$</span></p>
<p>we can set up a Lagrangian given by:</p>
<p><span class="math-container">$$\mathcal{L} = U(x,y) - \lambda( p_xx+ p_yy - I)$$</span></p>
<p>Solving the Lagrangian above we will get some optimal demands <span class="math-container">$x^*(I)$</span> and <span class="math-container">$y^*(I)$</span>. This will give us a value function:</p>
<p><span class="math-container">$$U^*(x^*(I),y^*(I))=U^*(I)$$</span></p>
<p>Assuming the value function is differentiable the differential of value function is given by:</p>
<p><span class="math-container">$$dU^*(I) = dU^*(x^*,y^*)= U_x'(x^*,y^*)dx^* +U_y'(x^*,y^*)dy^* \tag{*}  $$</span></p>
<p>No given the budget constraint which generally can be written as <span class="math-container">$g(x,y)=I$</span> where <span class="math-container">$g(x,y)=p_xx+ p_yy$</span>, the FOCs of the Lagrangian will be given by:</p>
<p><span class="math-container">$$U_x'(x^*,y^*)dx^* = \lambda g_x'(x^*,y^*)dx^* \text{ and } U_y'(x^*,y^*)dy^* = \lambda g_y'(x^*,y^*)dy^*$$</span></p>
<p>Hence * can be written as:</p>
<p><span class="math-container">$$dU^*(I) =\lambda g_x'(x^*,y^*)dx^*  +\lambda g_y'(x^*,y^*)dy^*  = \lambda \left( g_x'(x^*,y^*)dx^*  + g_y'(x^*,y^*)dy^*\right) \tag{**}$$</span></p>
<p>Now if we take the differential of the budget constraint we get:</p>
<p><span class="math-container">$$dg(x^*,y^*)= g_x'(x^*,y^*)dx^*  + g_y'(x^*,y^*)dy^*= dI$$</span></p>
<p>Substitute this into ** and you will get:</p>
<p><span class="math-container">$$dU^*(I) = \lambda dI \text{ or } \frac{dU^*(I)}{dI} = \lambda $$</span></p>
<p>The last expression states that change in value function <span class="math-container">$U^*$</span> will be equal to the change in income times lambda. We could also rewrite the last expression as $U^*(I+dI)\approx \lambda dI$$.</p>
<blockquote>
<p>How do I get a sense of this, both mathematically and intuitively?</p>
</blockquote>
<p>Mathematical derivation was provided above. Intuitively <span class="math-container">$\lambda$</span> parameter is the marginal effect of the constraints on optimal attainable value of utility.</p>

CC BY-SA 4.0
2022-07-02T10:32:03.047Z
Thank you very much. It addresses my exact question. Three quick follow-up questions if you don't mind. 1) Is this the Envelope theorem (considering you're using value function and all) that @ShaikhAmmar was referring to in the comments? 2) You showed that $dU^{*}(I) = \lambda dI$. Is it an approximation or an equality? When I tried out with a few examples, they seemed to be equal but then you have written $U^{*}(I+dI) \approx \lambda dI$ which confuses me a bit. 3) Does it work for larger changes as opposed to infinitesimal increase in income or utility?
cris
CC BY-SA 4.0
2022-07-02T10:36:40.930Z
You perhaps meant $U^{*}(I+dI) - U^{*}(I) \approx \lambda dI$. Please correct me if I am wrong.
cris
CC BY-SA 4.0
2022-07-02T10:44:42.680Z
@cris 1. it’s not direct use of envelope theorem, I used little bit roundabout way but envelope theorem gives the same result. 2. $dU^*(I)=\lambda dI$ is equality. When you replace $dU$ it becomes approximation since $dU^*(I) \approx U^*(I+dI)$. 3. It works for larger than infinitesimal changes as approximation that was point of that rewritten formula.
1muflon1
CC BY-SA 4.0
2022-07-02T10:01:21.800Z
2022-07-02T10:01:21.800Z
1muflon1
<p>I read that if the income increases by <span class="math-container">$\\\$d$</span>, then the utility at the optimal point will increase by <span class="math-container">$\lambda d$</span>. How do I get a sense of this, both mathematically and intuitively?</p>
<p>Can we write that as <span class="math-container">$\lambda = \frac{\Delta U}{\Delta I}$</span> where <span class="math-container">$U$</span> is the utility function and <span class="math-container">$I$</span> is the income? The preferences are assumed to be well-behaved and <span class="math-container">$p_xx + p_yy = I$</span> (respectively <span class="math-container">$I' = I+d$</span>) is the budget constraint.</p>

CC BY-SA 4.0
2022-07-02T07:56:25.540Z
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Giskard
CC BY-SA 4.0
2022-07-02T09:20:22.520Z
I guess you are asking about Envelope Theorem, you should look that up. KhanAcademy has a series on constrained optimisation, which covers that.
Shaikh Ammar
CC BY-SA 4.0
2022-07-02T09:46:26.320Z
@Giskard See [this](https://i.imgur.com/mKqMAtz.png). In the example given, $\Delta U = 0.02 = 0.25 \cdot 0.08 = \lambda \Delta I$. Does this hold in general?
cris
CC BY-SA 4.0
2022-07-02T17:08:43.717Z
In response to the edit currently being approved: the equality will only hold for constant-returns-to-scale functions (e.g., CRS Cobb-Douglas). Try it with $U(x,y) = \ln x + \ln y$.
Giskard
<microeconomics><demand><choice-theory>
If the income increases by $\$d$, will it mean the utility at the optimal point increases by $\lambda d$?
cris
# If the income increases by $\$d$, will it mean the utility at the optimal point increases by $\lambda d$?

(Asked by: username_0 on 2022-07-01)

I read that if the income increases by $\\\$d$, then the utility at the optimal point will increase by $\lambda d$. How do I get a sense of this, both mathematically and intuitively?

Can we write that as $\lambda = \frac{\Delta U}{\Delta I}$ where $U$ is the utility function and $I$ is the income? The preferences are assumed to be well-behaved and $p\_xx + p\_yy = I$ (respectively $I' = I+d$) is the budget constraint.

**Tags:** microeconomics, demand, choice-theory
> **Comments:**
> **username_1**: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.  
> **username_2**: I guess you are asking about Envelope Theorem, you should look that up. KhanAcademy has a series on constrained optimisation, which covers that.  
> **username_0**: @Giskard See [this](https://i.imgur.com/mKqMAtz.png). In the example given, $\Delta U = 0.02 = 0.25 \cdot 0.08 = \lambda \Delta I$. Does this hold in general?  
> **username_1**: In response to the edit currently being approved: the equality will only hold for constant-returns-to-scale functions (e.g., CRS Cobb-Douglas). Try it with $U(x,y) = \ln x + \ln y$.  

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## 1 answers

### Answer by username_3 on 2022-07-02. Score: 1

> If the income increases by $d, will it mean the utility at the optimal point increases by λd?

Yes it will. For a problem:

$$\max U(x,y) \text{ s.t. } p\_xx+ p\_yy=I$$

we can set up a Lagrangian given by:

$$\mathcal{L} = U(x,y) - \lambda( p\_xx+ p\_yy - I)$$

Solving the Lagrangian above we will get some optimal demands $x^\*(I)$ and $y^\*(I)$. This will give us a value function:

$$U^\*(x^\*(I),y^\*(I))=U^\*(I)$$

Assuming the value function is differentiable the differential of value function is given by:

$$dU^\*(I) = dU^\*(x^\*,y^\*)= U\_x'(x^\*,y^\*)dx^\* +U\_y'(x^\*,y^\*)dy^\* \tag{\*} $$

No given the budget constraint which generally can be written as $g(x,y)=I$ where $g(x,y)=p\_xx+ p\_yy$, the FOCs of the Lagrangian will be given by:

$$U\_x'(x^\*,y^\*)dx^\* = \lambda g\_x'(x^\*,y^\*)dx^\* \text{ and } U\_y'(x^\*,y^\*)dy^\* = \lambda g\_y'(x^\*,y^\*)dy^\*$$

Hence \* can be written as:

$$dU^\*(I) =\lambda g\_x'(x^\*,y^\*)dx^\* +\lambda g\_y'(x^\*,y^\*)dy^\* = \lambda \left( g\_x'(x^\*,y^\*)dx^\* + g\_y'(x^\*,y^\*)dy^\*\right) \tag{\*\*}$$

Now if we take the differential of the budget constraint we get:

$$dg(x^\*,y^\*)= g\_x'(x^\*,y^\*)dx^\* + g\_y'(x^\*,y^\*)dy^\*= dI$$

Substitute this into \*\* and you will get:

$$dU^\*(I) = \lambda dI \text{ or } \frac{dU^\*(I)}{dI} = \lambda $$

The last expression states that change in value function $U^\*$ will be equal to the change in income times lambda. We could also rewrite the last expression as $U^\*(I+dI)\approx \lambda dI$$.

> How do I get a sense of this, both mathematically and intuitively?

Mathematical derivation was provided above. Intuitively $\lambda$ parameter is the marginal effect of the constraints on optimal attainable value of utility.

> **Comments:**
> **username_0**: Thank you very much. It addresses my exact question. Three quick follow-up questions if you don't mind. 1) Is this the Envelope theorem (considering you're using value function and all) that @ShaikhAmmar was referring to in the comments? 2) You showed that $dU^{*}(I) = \lambda dI$. Is it an approximation or an equality? When I tried out with a few examples, they seemed to be equal but then you have written $U^{*}(I+dI) \approx \lambda dI$ which confuses me a bit. 3) Does it work for larger changes as opposed to infinitesimal increase in income or utility?  
> **username_0**: You perhaps meant $U^{*}(I+dI) - U^{*}(I) \approx \lambda dI$. Please correct me if I am wrong.  
> **username_3**: @cris 1. it’s not direct use of envelope theorem, I used little bit roundabout way but envelope theorem gives the same result. 2. $dU^*(I)=\lambda dI$ is equality. When you replace $dU$ it becomes approximation since $dU^*(I) \approx U^*(I+dI)$. 3. It works for larger than infinitesimal changes as approximation that was point of that rewritten formula.  

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