(1) (ds)/(dpm) = (ds)/(dpm)|u-constant

(1) \frac{ds}{dp_m} = \frac{ds}{dp_m}|_{u-constant} - m \frac{ds}{dl} (2) \frac{dm}{dp_s} = \frac{dm}{dp_s}|_{u-constant} - s \frac{dm}{dl} Where the first term of each equation is a substitition effect and the second term is an income effect. The symmetry of net substitution effects implies which of the following conditions regarding the two previous Slutsky equations? $\circ \frac{dm}{dp_s}|_{u-constant} < \frac{ds}{dp_m}|_{u-constant}$ $\circ \frac{dm}{dp_s}|_{u-constant} > \frac{ds}{dp_m}|_{u-constant}$ $\circ$ The symmetry of net substitution effects does not apply to these slutsky equations $\circ \frac{dm}{dp_s}|_{u-constant} = \frac{ds}{dp_m}|_{u-constant}$ True or False: If $\frac{ds}{dp_m} = \frac{dm}{dp_s}$, and $\frac{dm}{dp_s}|_{u-constant} = \frac{ds}{dp_m}|_{u-constant}$, then Slutsky equation (1) must be equal to Slutsky equation (2) and $s \cdot \frac{dm}{dl} = m \cdot \frac{ds}{dl}$ $\circ$ True $\circ$ False Use the Marshalian demand function you derived earlier to explicitly calculate the income effects, $m \frac{ds}{dl}$ and $s \frac{dm}{dl}$, from the previous slutsky equations. m \frac{ds}{dl} =

Transcript

00:01 So this is actually a lengthy problem.

00:02 I hope i can cover it fast enough.

00:05 Number one, in order to write out the utility max problem, you actually need one kind of constraint.

00:11 It's called a budget constraint.

00:13 So think of it like the price of product one times how much you're going to buy for the product one, which we call it x1, plus the price of x2, which product two, and then how much you're going to buy it.

00:28 The quantity is x1 x2 that will actually equal to your total budget which the problem called it w so in here remember that w is a constant so we're going to solve it right w is a constant and we want to maximize it in the way that you want the utility maximum with this budget constraint so the larangian equation l rangian which we call it l is actually going to be your utility function which is going to be x1 to the power alpha 1 x2 to the power 1 minus alpha plus a new constant called lambda lambda is also constant and then times your budget constraint which is w minus p1 x1 minus p2 x2 this is a method in calculus to find a maximum or the minimum and using the laurentian method so this is going to be your laurentian equation.

01:34 Part b is the interesting part.

01:36 That's how you actually find a maximum.

01:39 Well, in order to find a maximum, you actually need the three set of the ordinary differential equation.

01:45 That is, let me write it out, and then we're going to discuss how to find this.

01:49 That is the partial derivative of the laurentian with respect to x1, partial derivative with respect to x2 and then partial derivative with respect to lambda.

02:03 Over these, the alpha in this, in the original utility, alpha is also a constant.

02:12 So we're going to need to calculate these partial derivative, and then in order to solve it for part c, you're going to remember part c as you to solve it.

02:21 In order to solve it, you actually need to set the equation equal to zero.

02:24 So actually let's look at take partial derivative with x1, then x1 is variable that you're taking.

02:31 So you're going to have alpha 1, x1 to the power alpha 1 minus 1 as usual.

02:37 But you also have the x2 because that is not a variable that we're taking.

02:42 So we keep it the same.

02:44 And then plus lambda and omega is constant.

02:47 So derivative that would respect x1 to 0.

02:50 The only one that have x1 is the minus.

02:53 So i'm actually going to get it like this, get rid of that.

02:55 I'm going to write a minus here, minus the lambda p1, right? because that's the only term, the lambda and p1, x1 right there.

03:05 The other two is x2 and you take derivative that's going to be zero.

03:10 Okay.

03:10 In the same way, you're going to do this one.

03:13 We're doing derivative with respect to x2.

03:17 So you're going to have the coefficient is 1 minus alpha.

03:20 And then x2 is going to have the power is 1 minus alpha minus 1, which is just minus alpha, okay, to be, to be, let me write it out here, 1 minus alpha, and that is equal to minus alpha.

03:35 So you can see that.

03:37 And then don't forget the x1 term is still there because it's treated as a constant.

03:41 So alpha there.

03:44 There we go.

03:45 And then similarly, what do we got for this one? it's actually just minus lambda and the term go, what is p2 right there okay anything else is constant so derivative 0 and then lastly partial derivative is in respect to lambda the first term doesn't have any lambda so that is a zero only the second term is lambda so it's a whole thing multiply by lambda when you take derivative you get that whole thing so w minus p1 x1 minus p2 x2 and now the next step is to set each of these one equal to zero and let's name these number one number two number three if you look closely number three and if you set up to zero you're actually going to get your budget constrained because of you equal p1 x1 plus p2 x2 and there's nothing new about it so the tricky one is number one and number two so without turning page actually let me write it down but now careful i'm going to leave my alpha there's no one.

04:48 And then carefully rewrite this in term of power.

04:51 So you look at this is x1 power alpha minus 1 divided by x2 power alpha minus 1.

04:59 Can you see that? it's because this is x2 power 1 minus alpha which you can fact now it's going x1 power negative of alpha minus 1 and the negative is becoming the division right there.

05:12 You can see that and then that will equal to our of lambda p1.

05:19 It basically i set it equals zero so i can move this to the right hand side.

05:23 I can see that.

05:24 It's important to see this variety in this form...

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