Lesson 28-1: Direct and Indirect Effects in Total Derivatives
Suppose we have this situation where z is our dependent variable and z is a function of x and y.
= ( , )
Now in most instances, x and y are independent푧 푓 of푥 each푦 other. They’re not functionally related in most cases. But there are certain situations where y will be a function of x. So, suppose we had this: z is a function of x and y, and y is also a function of x. These situations occur sometimes where these two variables here are functionally related to each other.
In this type of set-up, you will have what’s called a total derivative of z with respect to x. Because, if you substitute in for y here, notice y is just a function of x.
= ( )
So, we could write this as f equals a function푦 of x,푔 g푥 of x.
= ( ) = ( , ( ))
So, in essence, g can be expressed as 푧a function푓 푥 of푓 x푥 alone.푔 푥 And you can find the derivative of z with respect to x.
But let me show how to do this in a slightly different way. Let’s take this expression here – z is a function of x and y – and let’s just forget about this part.
I don’t want to erase this. We’re still assuming y is a function of x.
Let’s take the total differential of z. The total differential of z will be exactly the same as before. What will the total differential be?
[Student comment]
Either partial z or partial f. Since we said z, I’ll put partial z.
[Student comment]
Generally, you’d do this one first because x comes first. Whatever comes first is what you’d generally do first.
= + 휕� 휕� 푑� 푑� 푑� [Student comment] 휕� 휕�
That’s the same expression. It’s still true whether x and y are functionally related or not. But, in this case here, we know that y is a function of x, so we could express z as a function of x.
So, we might be interested in dz/dx. In order to get dz/dx, the derivative of z with respect to x, notice this dz is a total differential. But if we want the derivative of z with respect to x, all we have to do is divide both sides of this equation by dx.
Doing so, we’d get dz/dx equals partial z partial x. Notice by dividing dx by dx, that’s just one. Plus partial z partial y dy/dx.
= + 푑� 휕� 휕� 푑� And this equation right here is called the푑 �total 휕derivative� 휕� 푑 �of z with respect to x. Notice this total derivative has two components. Partial z partial x, this term right here is referred to as the direct effect.
Direct effect:
휕� And clearly, it’s the direct effect of x on z, holding휕� y constant. It’s just the partial derivative. And that’s called the direct effect.
This term here is the indirect effect.
Indirect effect:
휕� 푑� It’s the indirect effect of x on z. Which is just휕 the� 푑 �effect of x on y multiplied times the effect of y on z. And it’s called the indirect effect. And this thing over here is called the total derivative.
Total derivative:
푑� 푑�
Lesson 28-2: Channel Map Illustration of the Total Derivative
= ( ) = ( , ( ))
푧 푓 푥 푓 푥 푔 푥 = + 푑� 휕� 휕� 푑� x is sort of the ultimate source of change푑 �in this휕 �model.휕� 푑 So,� notice x affects y. So, let’s draw an arrow: x affects y. That comes from this function right here: y is a function of x. So this arrow right here represents the effect of x on y. And I’ll just write: “via the g function.”
Now, y in turn affects z via the f function. The value of y affects the value of z through that f function. So, we’ll draw another arrow from y to z. So, I’ll say this arrow is via the f function.
And, there’s one more arrow here because x also has a direct effect on z via the f function. So, I have to draw one more arrow here, this one right here. That is the direct effect via the f function.
So, this is the direct effect. It’s the direct effect of x on z through the f function. And this set of two arrows right here is the indirect effect.
Does that make sense?
[Student comment]
The indirect effect is both of these arrows combined. Because, see, x affects y and y affects z. So, actually, mathematically, this indirect effect is the product of these two derivatives: dy/dx times partial z partial y. Mathematically, it’s the product of these two effects.
Lesson 28-3: Direct and Indirect Effects in Total Revenue
Let me give you a simple economic illustration of this.
And I’m going to take the simplest possible example I can think of.
Consider a firm’s revenue function. A firm that sells a single product. Revenue will be a function of price and output. Revenue depends on what the price per unit of the good is and the number of units you sell. And, in fact, this is a simple function. If the firm only produces one product, then its revenue is just P times Q.
= ( , )
푅 =푓 푃 푄
The quantity sold, Q, is going to be affected by푅 price.푃� So, you have Q is going to be some function g of price: = ( )
So, this fits exactly into the framework of the푄 total푔 derivative.푃 And, if we wanted to get the total derivative of revenue with respect to price, let’s again do it step by step.
What is dR in this case? The total differential.
[Student comment]
Let’s go back to the formula version first. We’ll go step by step.
[Student comment]
Partial R partial P times dP.
[Student comment]
= + 휕� 휕푅 푑� 푑� 푑� Now, given that the function f is just P times휕 �Q, what 휕is� partial R partial P? Q. Just Q.
If we take the derivative of PQ with respect to P, then it’s just Q. Q is the coefficient on P. So, it’s Q dP.
And what is partial R partial Q? Just P.
= 휕� 푄 휕� = 휕� 푃 So, the total differential of revenue is QdP plus휕� PdQ. Sounds kind of nice, doesn’t it? Sounds kind of funny. Q dP times P dQ.
Okay. Now, I said in this case, since Q is a function of P, we can get the total derivative of R with respect to P. And from this formula right here, how would we get dR dP?
= + 휕� 휕� 푑� 푑� 푑� [student comment] 휕� 휕�
Just divide both sides of this equation here by dP.
= +
푑� 푄𝑄 푃𝑃 = + 푑� 푑� 푑� 푄 푃 And you get dR/dP equals Q. dP/dP will푑� be what?푑� One.푑 � So, it will be just Q plus P times dQ/dP:
= + 푑� 푑� 푄 푃 And there’s your expression for the total 푑derivative� of푑 revenue� with respect to price. It’s just Q plus P times dQ/dP.
Lesson 28-4 Price Elasticity in the Total Revenue Function
= + 푑� 푑� 푄 푃 Actually, there’s something we can do with푑� this. Let’s푑 �play with this a little bit more.
Oftentimes, we’re interested in the sign of dR/dP, whether it’s zero, positive or negative. The firm would like to know this. If it raises price, what’ll happen to revenue? Will total revenue go up or will total revenue stay the same? Will total revenue go down? That would be a nice thing to know.
The answer to this question about what the sign of dR/dP is depends on the price elasticity of demand. And let’s show that.
We have this expression for dR/dP. dR/dP is equal to Q plus P times dQ/dP.
= + 푑� 푑� 푄 푃 Now this expression here is very close to푑 what?� 푑�
[Student comment] dQ/dP times P. If I wanted to convert that into the price elasticity of demand, what would I have to do?
[Student comment]
Just divide by, not P, Q. So, I can do this. I can divide by Q so long as I put another Q up here n the numerator. Because Q over Q is just one.
So, I can put a Q down here as long as I put another one up here in the numerator.
Then, this expression right here becomes Q plus E, the price elasticity of demand of Q with respect to P times Q.
= + = + , 푑� 푃 푑� 푄 푄 푄 퐸푄 푃푄 푑� 푄 푑� And now I can factor out my Q and rewrite this expression as Q times 1 plus the price elasticity of demand:
= + , = (1 + , ) 푑� 푄 퐸푄 푃푄 푄 퐸푄 푃 푑�
So, this formula is actually quite useful in microeconomics. If you know what the price elasticity of demand is, you could figure out whether dR/dP is positive, zero or negative.
For example, suppose the price elasticity of demand is exactly minus one. Then what is the sign of dR/dP?
If this price elasticity of demand right here is exactly minus one, what is the sign of dR/dP?
[Student comment]
Zero. Because you have one minus one, this is a zero times a Q is equal to dR/dP. So that means that dR/dP is equal to zero.
So, when the price elasticity of demand is exactly minus one, you raise price one percent, quantity demanded falls by 1 percent, your total revenue stays the same.
But, suppose the price elasticity of demand is greater than minus one.
Now greater than minus one means it’s minus point 9, minus point 8, minus point 7, something like that. Algebraically greater than minus one means it’s a smaller negative number.
In that case, what will the sign of dR/dP be? If this were, say, minus point 9?
[Student comment]
Yeah. One minus point 9 is point 1, so it would be positive.
So, the sign of dR/dP is going to be positive.
If the price elasticity of demand is less than minus one, then that means it’s minus 2, minus 3, etc. Then, what will the sign of dR/dP be?
[Student comment]
Negative. Because if this were minus 2 here, you’d have a one minus two would be minus one.
So, in fact, if it’s less than minus one, that means dR/dP is negative.
You probably learned this rule in introductory principles of microeconomics. And you probably learned the rule this way. When demand is elastic, meaning in absolute value greater than one, price and total revenue move in the opposite direction. If demand is elastic, price goes up, total revenue goes down. Price goes down, total revenue goes up.
If demand is inelastic, price and total revenue move in the same direction. Price goes up, total revenue goes up. Price goes down, total revenue goes down. If demand is inelastic.
If demand is unit elastic, meaning it’s exactly minus one, then total revenue does not change as price changes.
Do you recall this? Did you go over this in principles of micro? A rule like this?
Well, now you have a mathematical derivation of the rule, and you know where it comes from.
Let me just repeat. This case right here, demand is said to be unit elastic if it’s exactly one. In that case, when price changes, total revenue does not change at all.
In this case here, where the elasticity is greater than minus one, in algebraic value, means in absolute value, less than one. Demand is said to be inelastic.
And in this case, if the price elasticity is less than minus one, then demand is elastic.
Lesson 28-5: Total Derivative of Output with respect to Time
Suppose output is a function of K and L. And let this be an aggregate production function for an economy. An economy’s aggregate output depends on how much capital is used, how much labor is used.
= ( , )
And we’re going to study economic growth.푄 We’re푓 퐾 going퐿 to study how output changes over time.
So, over time, the capital stock will change. People will save. The savings will go into investment and it will cause the capital stock to change over time. So, let’s just let this function be k of t:
= ( )
This expression right here – instead of calling퐾 it f푘 of푡 t or g of t or h of t, let’s just call it k of t. And it’s understood capital here is a function of time.
And we’ll do the same thing for labor. Labor force will be a function of time. Perhaps just because of population growth, over time, the labor force will grow. The labor force could also grow if more people decide to work, if there’s an increase in labor force participation. But, as a simple explanation, you could just think of this as due to population growth. The population grows over time.
= ( )
Questions? 퐿 푙 푡
[Student comment]
It’s just a notation that is kind of convenient and is often used. I could – it would be all right – you could call this g of t and you could call this h of t but, sometimes, since it’s capital, they just write k parentheses t. And it’s understood that that means capital is a function of time. It’s just notation.
Actually, it’s kind of simplifying notation in a way because, if we used K equals g of t, then you have to think, “Oh, K and g are really the same thing.” But now you don’t have to because you just put a k instead of a g.
So, since K is a function of time and L is a function of time, notice that output can also be expressed a function of time. If I just substitute in here f of K, but K is a function of time, L is a function of time, so ultimately output is going to be a function of t.
= ( , ) = ( ( ), ( ))
푄 푓 퐾 퐿 푓 푘 푡 푙 푡
And we can get the total derivative of output with respect to time. It’s very easy. And again, let’s just start with dQ. And since you all know this by now, dQ is just partial Q partial K dK plus partial Q partial L dL. So, we’ll just start with that.
Notice Q is a function of K and L. In this case, notice t doesn’t appear in the f function. So, in this case, there’ll be just two indirect effects of time on output.
= + 휕� 휕� 푑� 푑� 푑� So, in order to get dQ dt, all we have to do is,휕� very simply,휕� divide both sides of this expression by dt.
= + 푑� 휕� 푑� 휕� 푑� And now, this thing right here is the total푑� derivative휕� 푑� of휕 output� 푑� with respect to time. I’ll just say “total derivative”. It’s obvious it’s the total derivative of Q with respect to t because dQ is in the numerator and dt is down here.
That’s the total derivative. And these are two indirect effects. Now you’ve got two indirect effects. Both of these are indirect effects of time on output. This one shows the effect of time on K and, in turn, the effect of K on Q. This indirect effect shows the effect of t on L and the effect of L on Q. So, both of these are indirect effects.
But this would be called the total derivative of output with respect to time.
There’s two components. As time changes, K changes and L changes. So, you’re going to have two indirect effects of t on Q.
Does that make sense to you?
Let’s draw the channel map for this problem. The channel map for this problem would look like this:
Here, our ultimate disturbance is t. When t changes, that causes K to change. Capital accumulates over time because of savings and then investment. And when K changes, that affects Q. So, that’s one indirect effect.
The other indirect effect is this: As t changes, L changes. And as L changes, Q changes.
So, in this particular model, you have two indirect effects of time on output. But no direct effect.
[Student comment]
That’s an excellent question.
Let’s do that. That would be a different model. But it actually is a model that is commonly used. So, let’s look at that.
Now, let’s just replace it with this:
= ( , , )
Q is a function of t comma K comma L. Now,푄 푓 t is푡 퐾in this퐿 f function. Again, it’s obvious that Q could be expressed as a function of time because you’ve got t in here. Then you’ve got K, but K itself is a function of time. And L is a function of time.
= ( , ( ), ( ))
So, now, if you want the total derivative푄 of Q푓 with푡 푘 푡respect푙 푡 to time, I would suggest you do it this way. Start off with the total differential, dQ. And dQ is just going to be partial Q partial t dt plus partial Q partial K dK plus partial Q partial L dL.
= + + 휕� 휕� 휕� 푑� 푑� 푑� 푑� That would be the total differential. Notice휕� now휕 there� are three휕� components. Because you’ve got three variables in here. But that’s the total differential of output.
Now, if you want dQ/dt, just divide both sides of this equation by dt and you get dQ/dt equals partial Q partial t. dt over dt is one. Plus partial Q partial K times dK/dt plus partial Q partial L times dL/dt.
= + + 푑� 휕푄 휕� 푑� 휕� 푑� Now this term right here is the direct푑� effect.휕� This휕� is푑 �the indirect휕� 푑� effect, same one we had before. And this is the indirect effect, too.
[Student comment]
I’m so proud of you people today. You’re really tying this stuff together. You’re asking good questions.
Yes, the only difference in this channel map, there would also be a direct effect right there. So, this would be the direct effect, this would be an indirect effect, and this is an indirect effect.
And, by the way, this model right here is a model that was used by Robert Solow in his study of economic growth in the US. I think he wrote this paper in 1957, a study of economic growth in the US. This is the framework he used.
And now time represented the state of technology. Over time, firms use better technology. And what that means is, for any given values of K and L, you get more output just because technology’s improving. Over time, K increases and L increases. And so, Solow used this type of model as a framework for studying economic growth in the US. And his work was very influential. And that’s in large part why he won the Noble Prize in economics, fifteen, twenty years ago, by using a model just like this to study the determinants of economic growth.
He not only had a theoretical model but then he empirically estimated it using data. And won the Nobel Prize. He used a model just like this.
Lesson 29-99: Review of Problem with Comparative Static Derivatives
We had the following supply and demand problem. Quantity demanded is given by this linear function – a minus bP plus e times I sub zero, where I sub zero is aggregate consumer income.
= + +
퐷 0 푄 >푎 0, 푏푏> 0,푒퐼 < 0
And notice we assume in this problem,푤ℎ푒푒푒 you’re푎 given푏 a is 푒positive, b is positive, and e is negative. So, the fact that e is negative means what kind of good is this? It means income has what kind of effect on quantity demanded?
[Student comment]
Yeah, it’s an inferior good because income has a negative effect on quantity demanded. As income goes up, quantity demanded goes down, so this is an inferior good in this case.
The supply function is just dP where d is a positive parameter:
= > 0
And, in equilibrium, of course, Qd equals푄푆 Qs.푑� 푤ℎ푒푒푒 푑
=
To solve for the equilibrium price, just set Qd푄 푑equal푄 푠to Qs and, doing so, we get:
+ =
Adding dP to both sides, we get: 푎 − 푏푏 푒퐼0 푑�
+ + = +
And now, factoring out the P on this푎 − right-hand푏푏 푏푏 side,푒퐼0 we푑 get:� 푏푏
+ = ( + )
Then, dividing both sides by (b plus d),푎 we 푒퐼get0 the 푏equilibrium푑 � price. And let me write the equilibrium price on the left-hand side rather than the right-hand side, so it will just be a plus e I zero over b plus d:
+ = ( + ) ∗ 푎 푒퐼0 Notice, on the right-hand side, all we have� are parameters and exogenous variable, so that’s the proper form. 푏 푑
So, now we’re ready to look at the comparative static derivatives. Now, what we will assume, we’ll assume that the equilibrium price is positive. If we assume that the equilibrium price is positive, what does that mean about a plus eI0 ? If the equilibrium price is positive, then what can we say about a plus eI0 ?
Well, what do we know about this denominator, b plus d?
[Student comment]
It’s positive. b is positive, d is positive. So, in order for the equilibrium price to be positive, what must be true about a plus eI0 ? Positive.
This is despite the fact that we know that e is negative. Okay. Even though e is negative and I is positive, it must be the case that a plus eI0 has to be positive in order for the equilibrium price to be positive. Okay. So, for this market to exist, the price, of course, has to be positive. Can’t have a negative price.
So, we’ll assume this as well.
Now, let’s do our comparative static derivatives.
Partial P star partial a is what?
[Student comment]
One over b plus d. And we know it has what sign? Positive. Because b and d are positive. So this has to be positive.
So, in words, that means P star and a move in the same direction.
1 = , > 0 ∗ + 휕� All right. Let’s do partial P star partial I.휕 � That’s푏 another푑 easy one. What’s that going to be?
[Student comment]
Consumer income is assumed to be exogenous in this model. So, what is partial P star partial I?
[Student comment]
The coefficient on I is just e over b plus d. And, given what we’ve assumed about e, what is the sign of this partial derivative? Negative. Because e is negative. So that implies what?
[Student comment]
I and P star move in the opposite direction.
= , < 0 ∗ + 휕� 푒 Then, I also asked for partial P star partial휕퐼 0b, didn’t푏 푑I? Actually, we know that partial P star partial b and partial P star partial d are going to be the same. So, you could verify this. These two will actually be the same. And what rule will we use to get partial P star partial b?
[Student comment]
And what would be if we used the quotient rule?
The derivative of the numerator with respect to be is zero. Times the denominator.
[Student comment]
The derivative of the denominator with respect to be is just going to be one. Times the numerator, all over the denominator squared.
0( + ) 1( + ) 1( + ) = = = = ∗ ∗ ( + ) ( + ) ( +∗ ) 휕� 휕� 푏 푑 − 푎 푒퐼0 푎 푒퐼0 � 2 2 − 2 So, this will be minus휕� a휕 plus� a plus eI0푏 over푑 b plus d squared.푏 푑 푏 푑
Or simplifying this, we could also write this as what?
[Student comment]
If you just take this, ignore the minus sign. This over one of these is just P star.
So this is just minus P star over b plus d.
So, the sign of these two partial derivatives is what? Negative. Meaning that b and P star move in the opposite direction. d and P star move in the opposite direction.
The last part of this problem was to show these effects using supply and demand diagrams.
Again, to repeat, our demand function is this, where we are assuming a positive, b positive, e negative.
= + +
퐷 0 푄 >푎 0, 푏푏> 0,푒퐼 < 0
푤ℎ푒푒푒 푎 푏 푒
So, and our supply curve is:
= > 0
Where d is assumed to be positive. So,푄푆 our푑 supply� 푤ℎ푒푒푒 and푑 demand diagram looks like this.
If we graph price on the horizontal axis and quantity on the vertical axis, how do we graph this demand curve? This is a little bit tricky.
What is the vertical intercept of the demand curve?
[Student comment]
Not quite a. Before, you’re right. If we didn’t have this term here, as we didn’t before, but now in this model, we’re adding this term.
So, if P is zero, what would the quantity demanded be?
It’s a plus eI. So, a plus eI is the intercept.
Now, we are assuming that a plus eI, as I explained earlier, is greater than zero, so it’s a positive intercept.
Then, the slope of the demand curve is just minus P. So, it’s the same as before except the intercept is now a plus eI.
So, the slope here is minus b.
And, since b is positive, the slope is negative.
The supply curve, if I draw the supply curve, what does it look like if we graph it?
[Student comment]
What does the supply curve look like?
[Student comment]
Because d is positive, right? What about the vertical intercept?
[Student comment]
Yeah, the origin. It comes right out of the origin.
So, the supply curve is just a straight line with a positive slope coming out of the origin. The slope of this is d.
Where they intersect, if you drop a perpendicular from this intersection right here, this would be your P star value. And let’s call this P star value zero for the initial situation. For the initial values of a, eI, and d.
All right. Now we can do comparative static analysis using our graph. If there’s an increase in a, what would that do to the demand curve? If a were to go up? What happens to the demand curve in this diagram?
[Student comment]
It shifts. It shifts vertically upward. The intercept increases, so it’ll shift. And what about the slope?
It’s not changing. Because the slope is minus b and b stays the same. So, the slope doesn’t change, so it’s a parallel shift upwards.
So, what happens to the equilibrium price?
It goes up. The new equilibrium point is right here. Let’s label the original equilibrium point A and label the new one B.
So, the new equilibrium price is P one star. That’s consistent with this result right here. P star and A move in the same direction.
You could illustrate each of the other shifts by just shifting the appropriate curve. In this case right here, if we were to illustrate I going up, what would change now?
Actually, I could use this same diagram. What would happen to the vertical intercept if I goes up?
Assume we’re starting with this case right here. This is the original demand curve. Here’s the new demand curve.
Well, I going up is also going to shift –
Oh, I made a mistake. What’s it going to do?
It’s going to shift it down. Because e is negative. Okay. e is negative.
So, we’re going to need a new diagram for this.
I was going too fast.
So, here’s our original diagram. a plus eI. Here’s our supply curve. Here’s our original equilibrium price.
Now, if I goes up, since e is negative, the new demand curve is going to shift downward, but the slope will be the same because minus b is still the slope. So, it’s a parallel shift.
So, here’s a plus eI zero prime, where I zero prime is greater than I zero.
So, now what happens to the price? It goes down. Which is consistent with our finding that this partial derivative is negative.
Well, I’ll let you do the other two. If b were to increase, what happens in the diagram if there’s an increase in b?
[Student comment]
A change in the slope of what?
[Student comment]
So, if b goes up, the demand curve then becomes steeper. The slope becomes more negative, so it becomes steeper. And that would obviously cause the equilibrium price to fall if the demand curve becomes steeper.