How was this equation differentiated?

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In summary, the conversation is about a mathematical problem involving a utility function and maximizing consumption subject to a budget constraint. The main point of confusion is the use of variables and the correct equation to use. The conversation also touches on the concept of a discount factor and the first order condition.
  • #1
amaya244
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Hi guys, this is my first post, a friend of mine said I should try the site out. Here it goes.I have a function: Max (c1) = u(c1) + 1/1+p * u(c2)

(c2) is equal to this: (1+r)^2 * A0 + (1+r) (Y1-c1) + Y2

Substituting it into the max gives: u(c1) + 1/1+p * u [(1+r)^2 * A0 + (1+r) (Y1-c1) + Y2]FOC, I got: u'(c1) = 1+r/1+p * u'(c2)

However, I think I'm wrong. Can someone please check this for me?

(please show working)

Amy xx
 
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  • #2
amaya244 said:
Hi guys, this is my first post, a friend of mine said I should try the site out. Here it goes.


I have an function: Max (c1) = u(c2) + 1/1+p * u(c2)

(c2) is equal to this: (1+r)^2 * A0 + (1+r) (Y1-C1) + Y2
I have no idea what you're asking. In the equation above, you have Max (c1) = u(c2) + 1/1+p * u(c2).

The way you wrote this suggests that Max is a function of variable named c1, but the other side of the equation involves c2, not c1.

Another point of confusion is trying to determine what you mean by u(c2) + 1/1+p * u(c2). Is u a number or a function? Also, what you wrote does not mean what you think it means. Apparently there is a fraction somewhere, but without parentheses, I can't tell what's in the numerator and what's in the denominator.

For example if you write (a + b)/(c + d) without parentheses, it would be a + b/c + d. Should this mean a + [b/(c + d)], [(a + b)/c] + d, or just plain a + [b/c] + d?
amaya244 said:
Substituting it into the max gives:


u(c2) + 1/1+p * u [(1+r)^2 * A0 + (1+r) (Y1-C1) + Y2]


FOC, I got: u'(c1) = 1+r/1+p * u'(c2)
What does FOC mean?
amaya244 said:
However, I think I'm wrong. Can someone please check this for me?

(please show working)

Amy xx
 
  • #3
Sorry Mark44, I totally understand what you mean. Let me explain a little.

Firstly, I made an error. c1 should indeed be in place of c2 (in the 1st utility function).

Secondly, 1/1+P is a 'discount factor' that can be simply written as a 'B':

u(c1) + B u(c2)

When B is close to 1, the individual is impatient and chooses to consume quite a lot of his future income.

Thirdly, let me explain the variables:

u(.) is an instantaneous utility function
ct is period t consumption
Yt is income at period t
At is wealth at period t

The problem is thus dynamic! The individual maximises consumption subject to a budget constraint.

u(c1) + B u(c2) s.t. c2 = (1+r)^2A0 + (1+r)(Y1-c1) + Y2

This is usually called Milton Friedman's Permanent Income Hypothesis. FOC is short for the 'first order condition'.

This first order condition is known as a Euler equation.
u'(c1) = (1+r) + B u'(c2)

I just want to know how c1 is maximised/differentiated.
 
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  • #4
amaya244 said:
Sorry Mark44, I totally understand what you mean. Let me explain a little.

Firstly, I made an error. c1 should indeed be in place of c2 (in the 1st utility function).

Secondly, 1/1+P is a 'discount factor' that can be simply written as a 'B':
That would be 1/(1 + P).
amaya244 said:
u(c1) + B u(c2)

When B is close to 1, the individual is impatient and chooses to consume quite a lot of his future income.

Thirdly, let me explain the variables:

u(.) is an instantaneous utility function
ct is period t consumption
Yt is income at period t
At is wealth at period t

The problem is thus dynamic! The individual maximises consumption subject to a budget constraint.

u(c1) + B u(c2) s.t. c2 = (1+r)^2A0 + (1+r)(Y1-c1) + Y2

This is usually called Milton Friedman's Permanent Income Hypothesis. FOC is short for the 'first order condition'.

This first order condition is known as a Euler equation.
u'(c1) = (1+r) + B u'(c2)

I just want to know how c1 is maximised/differentiated.

I think this is what you're trying to say:

Maximize u(c1) + B*u(c2), where c2 =A0(1 + r)2 + (1 + r)(Y1 - c1) + Y2.

If you substitute for c2 in the expression you want to maximize, you'll get a function that has one variable: c1.

Let's call this function F.
F(c1) = u(c1) + B*u(A0(1 + r)2 + (1 + r)(Y1 - c1) + Y2)).

The first part is easy enough to differentiate. For the second part you need to use the chain rule.
 
  • #5
Mark44 said:
That would be 1/(1 + P).

No, just 1/1 + P actually :-)


Mark44 said:
I think this is what you're trying to say:

Maximize u(c1) + B*u(c2), where c2 =A0(1 + r)2 + (1 + r)(Y1 - c1) + Y2.

If you substitute for c2 in the expression you want to maximize, you'll get a function that has one variable: c1.

Let's call this function F.
F(c1) = u(c1) + B*u(A0(1 + r)2 + (1 + r)(Y1 - c1) + Y2)).

The first part is easy enough to differentiate. For the second part you need to use the chain rule.

Thank you Mark44,

Can I just ask, once you differentiated the first and second parts of F, with respect to c1, did you arrive at the following equation?

u'(c1) = (1 + r) β*u'(c2)

Amy xx
 
  • #6
amaya244 said:
No, just 1/1 + P actually :-)
Yeah, right...
amaya244 said:
Thank you Mark44,

Can I just ask, once you differentiated the first and second parts of F, with respect to c1, did you arrive at the following equation?

u'(c1) = (1 + r) β*u'(c2)
Well, I didn't do the differentiation.

The equation you show here is the result of setting F'(c1) = 0, and then solving for u'(c1).
 
  • #7
Mark44 said:
Yeah, right...

Pffft!

Mark44 said:
Well, I didn't do the differentiation.

The equation you show here is the result of setting F'(c1) = 0, and then solving for u'(c1).

That's what I thought, but then where did C2 come from? Didn't it get substituted out?
 
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1. How do you differentiate an equation?

Differentiating an equation involves finding the derivative, which is a mathematical operation that calculates the rate of change of a function. To differentiate an equation, you need to apply differentiation rules, such as the power rule, product rule, or chain rule.

2. What is the purpose of differentiating an equation?

The purpose of differentiating an equation is to find the slope of the curve at a given point. This can be useful in many scientific fields, such as physics, economics, and engineering, as it allows us to analyze the behavior of a system over time and make predictions.

3. What is the difference between differentiation and integration?

Differentiation and integration are inverse operations. Differentiation finds the rate of change of a function, while integration finds the original function from its derivative. In other words, differentiation is like taking a photo of a moving object, while integration is like creating a video from a series of photos.

4. Can any equation be differentiated?

Yes, any equation can be differentiated as long as it is a continuous function. However, some equations may be more complicated to differentiate than others, and some may require the use of advanced techniques, such as implicit differentiation or logarithmic differentiation.

5. Are there any common mistakes to avoid when differentiating an equation?

Yes, there are a few common mistakes to avoid when differentiating an equation. These include forgetting to apply the chain rule, mixing up the order of terms when using the product rule, and not simplifying the final result. It is also important to carefully check your work for any algebraic errors.

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