Actuarial Outpost Question About Delta (Greeks)
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#1
02-15-2008, 04:10 PM
 Phantomwise Member Join Date: Apr 2006 Location: Dallas Posts: 46

Delta is the partial derivative of the option price with respect to the stock price and given to be calculated as $\Delta=e^{-\delta t}N(d_1)$.

Looking at Black-Scholes where $C=Se^{-\delta t}N(d_1)-Ke^{-rt}N(d_2)$, the given formula for Delta would be the partial of the option price with respect to the stock price, but only if $N(d_1)$ is a constant. But $d_1$ is dependent on stock price, so it's not a constant.

Can anyone help me understand why $\Delta=e^{-\delta t}N(d_1)$?
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Last edited by Phantomwise; 02-19-2008 at 11:15 AM.. Reason: Learning MimeTex
#2
02-15-2008, 07:45 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,311

The important point is that d2 also depends on S, and so
$\Delta = \frac{\partial C}{\partial S} = e^{- \delta t} N(d_1) + S e^{- \delta t} N'(d_1) \frac{\partial d_1}{\partial S} - e^{- r t} K N'(d_2) \frac{\partial d_2}{\partial S}$

But those last two terms exactly cancel out (you need to work out the partial derivatives of d1 and d2 for that to be clear, though).

EDIT: Oh, you probably also need to use the formula for $N'(x)$, too. It's been a little while since I checked this formula.
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Last edited by jraven; 02-15-2008 at 07:48 PM..
#3
02-15-2008, 08:02 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,311

Bah, I might as well just work it out.

We know $d_2 = d_1 - \sigma \sqrt{t}$, so that tells us that
$\frac{\partial d_2}{\partial S} = \frac{\partial d_1}{\partial S}$.

We also know that
$N'(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}$
so
$N'(d_2) = \frac{1}{\sqrt{2 \pi}} e^{- (d_1 - \sigma\sqrt{t})^2 / 2} = \frac{1}{\sqrt{2 \pi}} e^{- d_1^2 / 2}\, e^{d_1 \sigma \sqrt{t}}\, e^{- 0.5 \sigma^2 t}$

$= N'(d_1) e^{\ln(S/K) + (r - \delta + 0.5 \sigma^2) t}\, e^{- 0.5 \sigma^2 t} = N'(d_1) \left( \frac{S}{K} \right) e^{(r - \delta) t}$

Putting this into our formula for $\Delta$ gives
$\Delta = e^{-\delta t} N(d_1) + S e^{-\delta t} N'(d_1) \frac{\partial d_1}{\partial S} - K e^{-r t} \left[N'(d_1) \left( \frac{S}{K} \right) e^{(r - \delta) t} \right] \left[\frac{\partial d_1}{\partial S} \right]$

$= e^{-\delta t} N(d_1) + S e^{-\delta t} N'(d_1) \frac{\partial d_1}{\partial S} - S e^{-\delta t} N'(d_1) \frac{\partial d_1}{\partial S} = e^{-\delta t} N(d_1)$
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#4
02-19-2008, 11:10 AM
 Phantomwise Member Join Date: Apr 2006 Location: Dallas Posts: 46

I'd forgotten about $d_2$. It doesn't seem intuitive that the terms would cancel like that, but looking at your proof I see that it does indeed work out that way mathematically. Interesting.

It probably took you a while to type that out, but it does clear things up a great deal. Many thanks!
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#5
02-19-2008, 07:40 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,311

Quote:
 Originally Posted by Phantomwise It doesn't seem intuitive that the terms would cancel like that...
Well, there's a way of viewing Black-Scholes as a discounted expected value (which isn't part of the MFE material), and from that point of view it's much easier to see that the formula for Delta should be very simple.

However when all you have to work with is the final Black-Scholes formula there's not much you can do but take the derivative and grind away, and then it does seem like a bit of a miracle that the formula for Delta is so simple.
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#6
03-07-2008, 05:35 PM
 macTEN Member Join Date: Feb 2007 Posts: 205
SOA sample no. 18

(18) A market maker sells 1000 1-yr euro gap call options, and delta-hedges the position with shares. You are given:
1. each gap call is written on 1 share
2. S(0)=100
3.sigma=100%
4. K=130
5. trigger=100
6. r=0%
Under BS framework, determine the initial number of shares in the delta hedge.
ANS=586.

The solution is a direct application of the discussion above.
where C(gap)=S*N(d1)-130*N(d2) = [S*N(d1)-100*N(d2)]-30*N(d2)=C-30*N(d2)
Delta(gap)=partial derivative of C(gap) wrt S

Ive never thought about the N(d) formula into detail and exactly how it works with the BS formula.........can anyone shed some light on this

thanks
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#7
03-07-2008, 08:19 PM
 macTEN Member Join Date: Feb 2007 Posts: 205

Quote:
 Originally Posted by jraven Well, there's a way of viewing Black-Scholes as a discounted expected value (which isn't part of the MFE material), and from that point of view it's much easier to see that the formula for Delta should be very simple.
can anyone elaborate on this way of viewing BS. and also can this be used to solve SOA sample 18. (ie intuitively instead of the direct approach in the SOA solution)
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#8
03-07-2008, 09:52 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,311

Quote:
 Originally Posted by macTEN can anyone elaborate on this way of viewing BS. and also can this be used to solve SOA sample 18. (ie intuitively instead of the direct approach in the SOA solution)
I don't think it would really help with computing the delta of gap options; the problem is that a term which is negligible for ordinary options can't be ignored in the gap option case, and working out it's contribution is annoying. (Doable, but annoying.) The trick with writing the value of the gap option as a plain call plus an extra term is probably the simplest approach.
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The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#9
03-08-2008, 06:42 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,311

Quote:
 Originally Posted by jraven I don't think it would really help with computing the delta of gap options; the problem is that a term which is negligible for ordinary options can't be ignored in the gap option case, and working out it's contribution is annoying. (Doable, but annoying.) The trick with writing the value of the gap option as a plain call plus an extra term is probably the simplest approach.
This probably deserves a bit more explanation... The discounted expected value view of Black-Scholes basically says that
$C(S, K, t) = e^{-r t} E[(SX - K)_+]$
where $X$ is a lognormal random variable with parameters $\mu = (r - \delta - 0.5 \sigma^2) t$ and $\sigma = \sigma \sqrt{t}$ (where that second sigma is the volatility of the stock). The product $S X$ repesents the stock price at time t in a risk neutral world, so we're computing the discounted expected payoff in a risk-neutral world; but since the price of an option doesn't depend on the market's attitude about risk, this will give the correct price in the real world too. [Well, assuming stocks actually followed GBM in the real world.]

But then
$C(S, K, t) = e^{-r t} E[SX - K | X > K/S]\, P(X > K/S)$

$C(S, K, t) = e^{- r t} S\, E[X | X > K/S]\, P(X > K/S) - e^{-r t} K\, P(X > K/S)$

Some explicit computations for lognormal variables results in
$E[X | X > K/S] = e^{(r - \delta) t} \frac{N(d_1)}{N(d_2)}$

$P(X > K/S) = N(d_2)$

When those are plugged into the formula for the price of the option you get
$C(S, K, t) = e^{- \delta t} S N(d_1) - e^{- r t} K N(d_2)$

Ok, that's not very simple. But what does it tell us about $\Delta$? Well, informally speaking (because there's a HUGE catch hidden here)
$\Delta = \frac{\partial}{\partial S} e^{-r t} E[ (SX - K)_+ ] = e^{- rt} \frac{\partial}{\partial S} E[ (SX - K)_+ ] = e^{- rt} E\left[ \frac{\partial}{\partial S} (SX - K)_+ \right]$

But
$\frac{\partial}{\partial S} (SX - K)_+ = \left{\begin{array}{ll} X & \text{if}\, S > K / X \,(\text{i.e. if}\, X > K / S) \\ 0 & \text{if}\, S < K / X \,(\text{i.e. if}\, X < K / S) \end{array}\right.$

So
$\Delta = e^{- rt} E[ X | X > K / S] P(X > K / S) = e^{- \delta t} N(d_1)$
(using the formulas above for the expected value and probability).

So what's the catch? Can we actually move the derivative inside the expected value? The function $(SX - K)_+$ isn't differentiable with respect to $S$ when $S = K/X$ (something I specifically ignored when I gave the derivative above), which makes it questionable whether you can. It turns out that in this specific case you can move it inside the expectation, and everything I typed is fine. But in the case of a gap option, where
$\text{payoff} = \left{\begin{array}{ll} SX - L & \text{if}\, SX \geq K \\ 0 & \text{if}\, SX < K\end{array}\right.$

you can't (the fact that the payoff isn't continuous is a big problem). Anyway, it means you have to be more careful with computing the derivative of the expectation in the gap option case. And at that point you might as well just find $\Delta$ some other way.
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.

Last edited by jraven; 03-08-2008 at 07:11 PM..
#10
01-15-2009, 01:19 PM
 Marid Audran Join Date: Nov 2008 Location: Minnesota Studying for MFE/3F Posts: 14 Blog Entries: 17
Can we use the fact that Black-Scholes is homogeneous of degree 1 in K and S?

It looks like people were wondering about that Delta formula right around the same time a year ago

The expected value discussion is very interesting and seems essential...and yet I keep wondering whether there's a simpler derivation. A mathematician named Rolf Poulsen has a paper on the subject, in which he says this:
Euler's Theorem (that is otherwise predominantly used in microeconomics) says that a differentiable function f is homogenous [of degree 1] if and only if it has the form
$f(x_1, ..., x_n) = \sum_{i=1}^n x_i \frac {\partial f}{\partial x_i}$.
Now observe that the Black-Scholes call-price is homogenous in stock-price and strike. Then Euler’s Theorem tells us that the term that "multiplies S" in the formula is indeed the partial derivative with respect to S; the delta.
I have two problems with this: (1) I believe that there needs to be more discussion about the "term that multiplies S." A decomposition of a homogeneous function
$f(x,y) = xg(x,y) + yh(x,y)$
is not unique in general: For example, we have
$x(-\frac yx) + y.1 = 0$ and $x(-y^2/x) + y.y = 0$.
Therefore, you can't necessarily "read off" the partial derivatives of a homogeneous function from a decomposition like that. Hmm.

Also, (2) I'm used to spelling "homogeneous" with two e's

 Tags delta, delta hedging