http://www.speculativebubble.com/videos/real-estate-roller-coaster.php
Wednesday, April 25, 2007
Tuesday, April 24, 2007
Structured Products
Just got my hands on Structured Products by Satyajit Das. It consists of two volumes, 2600 pages total. Volume 1 covers applications of derivatives, synthetic assets, exotic options, and interest rate and currency structured products. Volume 2 covers equity linked structures, commodity linked structures, credit derivatives and new markets (e.g., inflation, insurance, weather, etc).
You'd think that such massive tomes, covering such an encyclopedic list of topics, contain a wealth of knowledge. You'd be wrong. The explanations are at such a basic level that they are essentially useless to anybody with even an elementary understanding of structured products. And to top it off, the material is incredibly dated (even though the third "revised" edition was published in 2006). For example, the author dedicates pages to employee stock option plans and how they are not reflected in the financial statements, with references to articles written in the 1990s. He appears completely unaware that IASB and FASB both now require employee stock options to be accounted for in company's financial statements.
A total waste of money.
You'd think that such massive tomes, covering such an encyclopedic list of topics, contain a wealth of knowledge. You'd be wrong. The explanations are at such a basic level that they are essentially useless to anybody with even an elementary understanding of structured products. And to top it off, the material is incredibly dated (even though the third "revised" edition was published in 2006). For example, the author dedicates pages to employee stock option plans and how they are not reflected in the financial statements, with references to articles written in the 1990s. He appears completely unaware that IASB and FASB both now require employee stock options to be accounted for in company's financial statements.
A total waste of money.
Saturday, April 21, 2007
WHD Problem 5.9
Calculate the theta, gamma, vega and rho for European call and put options.
Note: I have assumed a non-dividend stock.
Θ = ∂Π/∂t
Θc = - ½S0N'(d1)σ/√T - rKe-rTN(d2)
Θp = - ½S0N'(d1)σ/√T + rKe-rTN(-d2)
Γ = ∂2Π/∂S2
Γc = Γp = N'(d1)/S0σ√T
v = ∂Π/∂σ
vc = vp = S0N'(d1)√T
ρ = ∂Π/∂r
ρc = KTN(d2)e-rT
ρp = -KTN(-d2)e-rT
Note: I have assumed a non-dividend stock.
Θ = ∂Π/∂t
Θc = - ½S0N'(d1)σ/√T - rKe-rTN(d2)
Θp = - ½S0N'(d1)σ/√T + rKe-rTN(-d2)
Γ = ∂2Π/∂S2
Γc = Γp = N'(d1)/S0σ√T
v = ∂Π/∂σ
vc = vp = S0N'(d1)√T
ρ = ∂Π/∂r
ρc = KTN(d2)e-rT
ρp = -KTN(-d2)e-rT
Thursday, April 19, 2007
Transforming the Black-Scholes equation into the Heat equation
(∂V/∂t) + ½ σ2 S2 (∂2V/∂S2) + r S (∂V/∂S) - rV = 0
First substitution: u = V e-rt
V = u ert
(∂u/∂t) = (∂V/∂t) e-rt - V r e-rt
(∂u/∂t) + V r e-rt = (∂V/∂t) e-rt
(∂V/∂t) = ert (∂u/∂t) + V r = ert (∂u/∂t) + r u ert
(∂u/∂S) = e-rt (∂V/∂S)
(∂V/∂S) = ert (∂u/∂S)
(∂2V/∂S2) = ert (∂2u/∂S2)
And the equation changes to...
ert (∂u/∂t) + r u ert + ½ σ2 S2 ert (∂2u/∂S2) + r S ert (∂u/∂S) - r ert u = 0
(∂u/∂t) + ½ σ2 S2 (∂2u/∂S2) + r S (∂u/∂S) = 0
Second substitution: S = ex
x = ln S
(∂S/∂x) = ex
(∂x/∂S) = 1 / S
(∂u/∂x) = (∂u/∂S) (∂S/∂x) = (∂u/∂S) ex = S (∂u/∂S)
S2 (∂2u/∂S2) = S2 ∂/∂S (∂u/∂S)
= S2 (∂x/∂S) ∂/∂x (∂u/∂S)
= S2 (1/S) ∂/∂x (1/S ∂u/∂x)
= S [ 1/S ∂2u/∂x2 + ∂u/∂x ∂/∂x (1/S) ]
= ∂2u/∂x2 + S (∂u/∂x) (-1/S2 ∂S/∂x)
= ∂2u/∂x2 - (1/S) ex (∂u/∂x)
= ∂2u/∂x2 - ∂u/∂x
And the equation changes to...
(∂u/∂t) + ½ σ2 (∂2u/∂x2 - ∂u/∂x) + r (∂u/∂x) = 0
(∂u/∂t) + ½ σ2 ∂2u/∂x2 + (r - ½σ2) ∂u/∂x = 0
Third substitution: z = x - (r - ½σ2)t to cancel the first derivative term, and t' = - t to conform to the usual sign convention.
(∂u/∂t) = (∂u/∂z)(∂z/∂t) + (∂u/∂t')(∂t'/∂t) = (∂u/∂z)[-(r - ½σ2)] + (∂u/∂t')(-1)
∂u/∂x = ∂u/∂z
∂2u/∂x2 = ∂2u/∂z2
And the equation changes to (dropping the ' on the t variable) ...
- (∂u/∂t) - (r - ½σ2) (∂u/∂z) + ½σ2 (∂2u/∂z2) + (r - ½σ2) (∂u/∂z) = 0
- (∂u/∂t) + ½σ2 (∂2u/∂z2) = 0
And finally, voila, we have the heat equation...
∂u/∂t = ½σ2 (∂2u/∂z2)
First substitution: u = V e-rt
V = u ert
(∂u/∂t) = (∂V/∂t) e-rt - V r e-rt
(∂u/∂t) + V r e-rt = (∂V/∂t) e-rt
(∂V/∂t) = ert (∂u/∂t) + V r = ert (∂u/∂t) + r u ert
(∂u/∂S) = e-rt (∂V/∂S)
(∂V/∂S) = ert (∂u/∂S)
(∂2V/∂S2) = ert (∂2u/∂S2)
And the equation changes to...
ert (∂u/∂t) + r u ert + ½ σ2 S2 ert (∂2u/∂S2) + r S ert (∂u/∂S) - r ert u = 0
(∂u/∂t) + ½ σ2 S2 (∂2u/∂S2) + r S (∂u/∂S) = 0
Second substitution: S = ex
x = ln S
(∂S/∂x) = ex
(∂x/∂S) = 1 / S
(∂u/∂x) = (∂u/∂S) (∂S/∂x) = (∂u/∂S) ex = S (∂u/∂S)
S2 (∂2u/∂S2) = S2 ∂/∂S (∂u/∂S)
= S2 (∂x/∂S) ∂/∂x (∂u/∂S)
= S2 (1/S) ∂/∂x (1/S ∂u/∂x)
= S [ 1/S ∂2u/∂x2 + ∂u/∂x ∂/∂x (1/S) ]
= ∂2u/∂x2 + S (∂u/∂x) (-1/S2 ∂S/∂x)
= ∂2u/∂x2 - (1/S) ex (∂u/∂x)
= ∂2u/∂x2 - ∂u/∂x
And the equation changes to...
(∂u/∂t) + ½ σ2 (∂2u/∂x2 - ∂u/∂x) + r (∂u/∂x) = 0
(∂u/∂t) + ½ σ2 ∂2u/∂x2 + (r - ½σ2) ∂u/∂x = 0
Third substitution: z = x - (r - ½σ2)t to cancel the first derivative term, and t' = - t to conform to the usual sign convention.
(∂u/∂t) = (∂u/∂z)(∂z/∂t) + (∂u/∂t')(∂t'/∂t) = (∂u/∂z)[-(r - ½σ2)] + (∂u/∂t')(-1)
∂u/∂x = ∂u/∂z
∂2u/∂x2 = ∂2u/∂z2
And the equation changes to (dropping the ' on the t variable) ...
- (∂u/∂t) - (r - ½σ2) (∂u/∂z) + ½σ2 (∂2u/∂z2) + (r - ½σ2) (∂u/∂z) = 0
- (∂u/∂t) + ½σ2 (∂2u/∂z2) = 0
And finally, voila, we have the heat equation...
∂u/∂t = ½σ2 (∂2u/∂z2)
WHD Problem 3.6b
Find the most general solution of the Black-Scholes equation that has the special form V=A(t)B(S).
(∂V/∂t) + ½ σ2 S2 (∂2V/∂S2) + r S (∂V/∂S) - rV = 0
V = A(t) B(s)
(∂V/∂t) = B (dA/dt)
(∂V/∂S) = A (dB/dS)
(∂2V/∂S2) = A (d2B/dS2)
B (dA/dt) + ½ σ2 S2 A (d2B/dS2) + r S A (dB/dS) - r A B = 0
(1/A) (dA/dt) + ½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = 0
½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = - (1/A) (dA/dt)
Since the left-hand side depends only on S and the right-hand side depends only on t, the only way the equality can hold is if both sides are equal to a constant K.
- (1/A) (dA/dt) = K
A(t) = c e-Kt
½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = K
½ σ2 S2 (d2B/dS2) + r S (dB/dS) - (r + K) B = 0
S2 (d2B/dS2) + (2r/σ2) S (dB/dS) - (2/σ2) (r+K) B = 0
This equation proceeds as in the previous exercise.
λ2 + (2r/σ2 -1)λ + (-2/σ2)(r+K) = 0
λ = { (1 - 2r/σ2) ± [ (2r/σ2 - 1)2 + (8/σ2)(r+K) ]½ } / 2
λ = { (1 - 2r/σ2) ± [ (2r/σ2 - 1)2 + (8/σ2)(r+K) ]½ } / 2
λ = { (1 - 2r/σ2) ± [ (2r/σ2 + 1)2 + (8K/σ2) ]½ } / 2
There are three cases depending on the roots: two real roots, one real root, two complex roots. And to be explicit, we should note that the case of two real roots can come about either with a real value of K (which leaves the form of A(t) above unchanged) or with a complex value of K (which would require rewriting A as we'll see in a minute).
Case 1: λ1 and λ2 are distinct real roots.
B(S) = c1 Sλ1 + c2 Sλ2
V(S,t) = (c1 Sλ1 + c2 Sλ2) e-Kt
Case 1A: K = 0
(Special scenario under case 1)
This reduces to the special case where there is no time dependence V = B(S)
V = c1 S + c2 S (-2r/σ2)
Case 2: λ1 = λ2 = λ is the only real root.
B(S) = Sλ (c1 + c2 ln S)
V(S,t) = Sλ (c1 + c2 ln S) e-Kt
Case 3: λ1 = a+ib and λ2 = a-ib are complex roots (with K real).
B(S) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ]
V(S,t) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ] e-Kt
Case 3A: λ1 = a+ib and λ2 = a-ib are complex roots (with complex K = c + id).
(This is not strictly separate from case 3; it's simply a matter of expanding e-Kt for complex K.)
B(S) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ]
V(S,t) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ] e-ct (cos dt - i sin dt)
Note that all of these solutions (except Case 1A) have too many degrees of freedom, due to the freedom of choice in choosing K. Normally boundary conditions would restrict the universe of valid values for K, but the problem as stated did not provide any such conditions.
(∂V/∂t) + ½ σ2 S2 (∂2V/∂S2) + r S (∂V/∂S) - rV = 0
V = A(t) B(s)
(∂V/∂t) = B (dA/dt)
(∂V/∂S) = A (dB/dS)
(∂2V/∂S2) = A (d2B/dS2)
B (dA/dt) + ½ σ2 S2 A (d2B/dS2) + r S A (dB/dS) - r A B = 0
(1/A) (dA/dt) + ½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = 0
½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = - (1/A) (dA/dt)
Since the left-hand side depends only on S and the right-hand side depends only on t, the only way the equality can hold is if both sides are equal to a constant K.
- (1/A) (dA/dt) = K
A(t) = c e-Kt
½ σ2 S2 (1/B) (d2B/dS2) + r S (1/B) (dB/dS) - r = K
½ σ2 S2 (d2B/dS2) + r S (dB/dS) - (r + K) B = 0
S2 (d2B/dS2) + (2r/σ2) S (dB/dS) - (2/σ2) (r+K) B = 0
This equation proceeds as in the previous exercise.
λ2 + (2r/σ2 -1)λ + (-2/σ2)(r+K) = 0
λ = { (1 - 2r/σ2) ± [ (2r/σ2 - 1)2 + (8/σ2)(r+K) ]½ } / 2
λ = { (1 - 2r/σ2) ± [ (2r/σ2 - 1)2 + (8/σ2)(r+K) ]½ } / 2
λ = { (1 - 2r/σ2) ± [ (2r/σ2 + 1)2 + (8K/σ2) ]½ } / 2
There are three cases depending on the roots: two real roots, one real root, two complex roots. And to be explicit, we should note that the case of two real roots can come about either with a real value of K (which leaves the form of A(t) above unchanged) or with a complex value of K (which would require rewriting A as we'll see in a minute).
Case 1: λ1 and λ2 are distinct real roots.
B(S) = c1 Sλ1 + c2 Sλ2
V(S,t) = (c1 Sλ1 + c2 Sλ2) e-Kt
Case 1A: K = 0
(Special scenario under case 1)
This reduces to the special case where there is no time dependence V = B(S)
V = c1 S + c2 S (-2r/σ2)
Case 2: λ1 = λ2 = λ is the only real root.
B(S) = Sλ (c1 + c2 ln S)
V(S,t) = Sλ (c1 + c2 ln S) e-Kt
Case 3: λ1 = a+ib and λ2 = a-ib are complex roots (with K real).
B(S) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ]
V(S,t) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ] e-Kt
Case 3A: λ1 = a+ib and λ2 = a-ib are complex roots (with complex K = c + id).
(This is not strictly separate from case 3; it's simply a matter of expanding e-Kt for complex K.)
B(S) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ]
V(S,t) = Sa [ c1 cos (b ln S) + c2 sin (b ln S) ] e-ct (cos dt - i sin dt)
Note that all of these solutions (except Case 1A) have too many degrees of freedom, due to the freedom of choice in choosing K. Normally boundary conditions would restrict the universe of valid values for K, but the problem as stated did not provide any such conditions.
Wednesday, April 18, 2007
WHD Problem 3.6a
Find the most general solution of the Black-Scholes equation that has the special form V=V(S).
(∂V/∂t) + ½ σ2 S2 (∂2V/∂S2) + r S (∂V/∂S) - rV = 0
Substituting V=V(S) reduces the partial differential equation to an ordinary differential equation.
S2 (d2V/dS2) + (2r/σ2) S (dV/dS) - (2r/σ2) V = 0
This is an Euler differential equation x2y'' + axy' - by = 0, the solutions of which can be found with the aid of the characteristic equation.
λ2 + (a-1)λ + b = 0
λ2 + (2r/σ2 -1)λ + (-2r/σ2) = 0
Applying the quadratic equation, one obtains (after a bit of algebra) two roots
λ = 1 and λ = (-2r/σ2)
Using these roots, we determine that the most general solution of the Euler equation is
V(S) = c1 S + c2 S (-2r/σ2)
(∂V/∂t) + ½ σ2 S2 (∂2V/∂S2) + r S (∂V/∂S) - rV = 0
Substituting V=V(S) reduces the partial differential equation to an ordinary differential equation.
S2 (d2V/dS2) + (2r/σ2) S (dV/dS) - (2r/σ2) V = 0
This is an Euler differential equation x2y'' + axy' - by = 0, the solutions of which can be found with the aid of the characteristic equation.
λ2 + (a-1)λ + b = 0
λ2 + (2r/σ2 -1)λ + (-2r/σ2) = 0
Applying the quadratic equation, one obtains (after a bit of algebra) two roots
λ = 1 and λ = (-2r/σ2)
Using these roots, we determine that the most general solution of the Euler equation is
V(S) = c1 S + c2 S (-2r/σ2)
Saturday, April 14, 2007
WHD Problem 2.5
There are n assets Si,...,Sn satisfying the following stochastic differential equations dSi = σi Si dXi + μi Si dt, for i = 1, ..., n
The Wiener processes dX i satisfy E[dXi] = 0 and E[dXi2] = dt, as usual, but the asset price changes are correlated with E[dXidXj] = ρijdt where -1 ≤ ρij ≤ 1.
Derive Ito's Lemma for a function f(Si,...,Sn) of the n assets.
Actually, I already know most or all of the content of The Mathematics of Financial Derivatives; I am reading it as a review in preparation for more advanced materials I am hoping to get to in the near future.
This exercise is pretty trivial, mostly a matter of keeping track of all the variables.
The Taylor series expansion is
df = (∂f/∂t)dt + Σ(∂f/∂Si)dSi + ½Σ(∂2f/∂Si∂Sj)dSidSj
dSidSj = (σi Si dXi + μi Si dt)(σj Sj dXj + μj Sj dt) = σi σj Si Sj dXi dXj = ρij σi σj Si Sj dt
dropping all higher order terms and then substituting in the correlation assumption.
So finally we obtain
df = [ (∂f/∂t) + Σ μi Si (∂f/∂Si) + ½ Σ ρij σi σj Si Sj (∂2f/∂Si∂Sj) ] dt + Σ σi Si (∂f/∂Si) dXi
The Wiener processes dX i satisfy E[dXi] = 0 and E[dXi2] = dt, as usual, but the asset price changes are correlated with E[dXidXj] = ρijdt where -1 ≤ ρij ≤ 1.
Derive Ito's Lemma for a function f(Si,...,Sn) of the n assets.
Actually, I already know most or all of the content of The Mathematics of Financial Derivatives; I am reading it as a review in preparation for more advanced materials I am hoping to get to in the near future.
This exercise is pretty trivial, mostly a matter of keeping track of all the variables.
The Taylor series expansion is
df = (∂f/∂t)dt + Σ(∂f/∂Si)dSi + ½Σ(∂2f/∂Si∂Sj)dSidSj
dSidSj = (σi Si dXi + μi Si dt)(σj Sj dXj + μj Sj dt) = σi σj Si Sj dXi dXj = ρij σi σj Si Sj dt
dropping all higher order terms and then substituting in the correlation assumption.
So finally we obtain
df = [ (∂f/∂t) + Σ μi Si (∂f/∂Si) + ½ Σ ρij σi σj Si Sj (∂2f/∂Si∂Sj) ] dt + Σ σi Si (∂f/∂Si) dXi
Thursday, April 12, 2007
Here we go...
So, let me get this straight those of us who did the RESPONSIBLE thing and did not buy more house than we could afford (while watching prices skyrocket out of our reach due to these irresponsible buyers) are now being asked to bend over and take it once more as our tax money goes to subsidize those who did buy more house than they could afford? Why?
And I'm tired of hearing that folks didn't understand what they were getting as a mortgage. Everyone understands adjustable interest. And everyone understands whether or not a payment is more than they can afford. And last of all everyone should understand the consequences of lying about his/her income on a mortgage application. I don't have any sympathy.
These folks took a risk. If it had worked out for them, the profit was all for them. They weren't planning to share it with me. It didn't work out, so I have to share the loss? WTF?
Tuesday, April 10, 2007
BearingPoint
The Company's continuing failure to timely file certain required periodic reports with the SEC imposes significant risks to the Company's business, including the possible loss of business, delisting from the New York Stock Exchange and defaults under the Company's credit facility. The Company has identified material weaknesses in its internal control over financial reporting, which could materially and adversely affect its business and financial condition. The Company's current cash resources might not be sufficient to meet its expected near-term cash needs, especially to fund intra-quarter operating cash requirements and non-recurring cash requirements (e.g., to settle lawsuits).
Apparently, BearingPoint didn't file its annual reports for fiscal 2004 and 2005 on time, nor will it file its 2006 report on time. It also has failed to file quarterly reports on time for the past six quarters in a row.
Apparently, BearingPoint didn't file its annual reports for fiscal 2004 and 2005 on time, nor will it file its 2006 report on time. It also has failed to file quarterly reports on time for the past six quarters in a row.
Friday, April 06, 2007
MBA Salaries
Came across the following in the WSJ...
The following business schools reported the highest average annual base salaries for full-time graduates, according to "The Wall Street Journal Guide to the Top Business Schools 2006".Stanford University: $100,400 Harvard University: $99,848 Massachusetts Institute of Technology: $94,131 University of Pennsylvania: $92,986 Dartmouth College: $91,900 Northwestern University: $91,390
The following business schools reported the highest average annual base salaries for full-time graduates, according to "The Wall Street Journal Guide to the Top Business Schools 2006".
Monday, April 02, 2007
LSI Swallowed Agere Today
Agere (a spin-off from Lucent) was acquired by LSI, effective today. The purchase price was 2.16 shares of LCI ($22.55) for each Agere share.
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