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June 21

Differentiation

I'm looking at the assumed knowledge for a financial mathematics course, and I'm ok with all of it except this one question:

Differential Calculus

As an indication of the level of calculus required, students should be able to interpret the following equation:

D={-}{(1+y) \over P}{dP \over dy}

I only know very basic differentiation (e.g. polynomials, basic trig functions). I've spent some time trying to figure it out but it's got me stumped. I can't tell if P is meant to be a function of y, or if D is meant to be a constant... I always thought capital letters in calculus just referred to the integral of an expression.

I understand if you don't want to tell me the answer, but can someone please identify the specific area of differential calculus I should be researching? And even better, a website/chapter in a book that can help me with this?

Thanks,

60.242.124.184 (talk) 12:19, 21 June 2008 (UTC)[reply]

P has to be a function of y, or differentiating it with respect to y is meaningless. D could be a constant or a function of y. If I had to guess, I think I'd go for a constant, since that's a common use for capitals from the beginning of the alphabet (though one normally starts with either A or C). Algebraist 12:27, 21 June 2008 (UTC)[reply]

This looks like a formula for bond duration (that is a really poor article). P is the price of the bond and y is the yield to maturity. So P is definitely a function of y. For small interest rate moves, D is typically treated as a constant but, in theory, it will change if the other variables change.

Duration is a measure of the sensitivity of the price of a fixed income asset to changes in interest rates. The Macaulay Duration is calculated as the average time to each cashflow weighted by the PV of that cashflow. Express the value as

P=\sum _{i}PV(CF_{i})=\sum _{i}CF_{i}(1+y)^{-t_{i}}

and find the derivative wrt y to see how the formula above gives the Macaulay Duration. Zain Ebrahim (talk) 13:25, 21 June 2008 (UTC)[reply]

Given that we are expected to use dP/dy I'd automatically assume that P is a function of y, and (again) in the absence of further information that D is a constant - since the alternatives (eg D is also a function of y) are just too complex..

So I'd rearrange to get

dP/dy = -D P/(1+y)

ie

d fn(y)/dy = k fn(y)/(1+y) (k is a constant equal to -D)

SPOILER BELOW ?? to avoid just read the bolded and underlined bits first

Since there is a equation that has the derivative of a function to be the function times another function of y , my first guess would be to try fn(y)=e^g(y) since

d(e^g(y))/dy = g'(y) e^g(y)

ie if P=e^g(y) then dP/dy = g'(y) P (g'(y) is the derivative of g(y) = d g(y)/dy )

Therefor I get one solution if g'(y) = k/(1+y), and because I already know that d (ln(1+x))/dx = 1/(1+x) I get g(y)=k ln(1+y)

So I think g(y)=k ln(1+y)

This gives P=e^{k ln(1+y)} = (e^ln(1+y))^k = (1+y)^k

So P=(1+y)^-D

Hopefully that will help, please ask about any specific parts, and hppefully others will pick me up on any mistakes I've made.

That should be one type of solution - I've no idea if there may be others as well...

As general advice - if you get a differential equation of the form d fn(x)/dx = fn(x) times another function of x it's worth trying the substituion fn(x)=e^g(x) due the properties of the derivative of e^g(x)87.102.86.73 (talk) 14:35, 21 June 2008 (UTC)[reply]

The thing is, it didn't say to solve it, it said to interpret it. That could mean any number of things, but probably nothing as complicated as actually solving for P. --Tango (talk) 14:57, 21 June 2008 (UTC)[reply]

The area to look at would be "first order differential equations", assuming you know how to differentiate (and integrate) polynomials, logs, and exponentials the equation should be ok..

Even if D is a function of y it may be possible to still solve it with the above knowledge.

I assumed that an answer would be a simple way to show the type of knowledge required..

In general capital letters don't mean the integral of an expression. They simply refer to a variable (which may or may not be a function) in the same way lower case letters do.87.102.86.73 (talk) 15:14, 21 June 2008 (UTC)[reply]

Indeed, however D(x) is often used for the integral of d(x), which is probably what the OP was thinking. I would never assume that's what was meant unless it was actually stated, though - it's a common choice of symbol, it's not a standard notation. --Tango (talk) 15:17, 21 June 2008 (UTC)[reply]

To the original poster: In financial mathematics, the formula you wrote is a very common way to express Macaulay Duration. You should check with the course convenor at the institution (they're normally quite responsive to emails) whether you need to know how to solve that differential equation but I rather doubt it since they clearly specified that you need to interpret it. The solution calculated by 87 above represents a special case: here, P represents the price of a zero coupon bond (ZCB) that pays one dollar D years from now. For ZCBs, the duration is merely the term to maturity. Zain Ebrahim (talk) 15:27, 21 June 2008 (UTC)[reply]

Thanks a lot everyone 60.242.124.184 (talk) 02:19, 22 June 2008 (UTC)[reply]