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01 MFE Basics

  • SSP 

Put-Call Parity Equation

$C-P = F_{0,T}^P (S)- F_{0,T}^P(K)$

$max[0,S-K] – max[0,K-S] = S-K$


Binomial Tree Model

$u = e^{(r-\delta)h+\sigma\sqrt h}$  | $d = e^{(r-\delta)h-\sigma\sqrt h}$

$p^* = \frac{e^{(r-\delta)h}-d}{u-d}$ [btw] = $ \frac{1}{1+e^{\sigma \sqrt h}} $[/btw]

$C_u = e^{-rh} [p^* .C_{uu} + (1-p^*).C_{ud}]$ | $C_d = e^{-rh} [p^* .C_{ud} + (1-p^*).C_{dd}]$

$ \Delta = e^{-\delta h}\frac{C_u – C_d}{S_0 (u-d)} $ | $ B = e^{-rh}\frac{uC_d -d C_u}{ u-d}$

$\text{Option Price} = e^{-r(2h)}[\binom{2}{2}p^{*^2}C_{uu} + \binom{2}{1}p^{*}(1-p^*)C_{ud} + \binom{2}{0}(1-p^*)^2C_{dd}]$

No dividends => American call option Price = European call option price. This is because, early exercise is never optimal.

Exchange Options

C(A,B) -> C(Receive A, Give up B)

P(A,B) -> P(Give up A, Receive B)

$C(A,B) – P(A,B) = F_{t,T}^P(A) – F_{t,T}^P(B)$

$C(A,B) = P(B,A)$

 

European Exchange option pricing

$ F^P(S)= Se^{-\delta_{s} (T-t)} $  |  $ F^P(Q) = Qe^{-\delta_{q} (T-t)} $  |  $ v^2 = \sigma ^2 (T-t) $  |  $ \sigma ^2 = \sigma_s^2 + \sigma_q^2 – 2\rho\sigma_s\sigma_q$

$ d_{1} = \frac{ln(\frac{F^P(S)}{F^P(Q)}) + [0.5*v^2]}{v}$
$ d_{2} = d_{1} – v$

[icon name=”bell” class=””] Price of a call to exchange stock S for stock Q. Payout at expiration = $(S_T – Q_T)_+$ :

Call Price $ =   [ F^P(S) * N(  d_{1})]  – [ F^P(Q) * N( d_{2})]$

Put Price $ = – [ F^P(S) * N(-d_{1})] + [ F^P(Q) * N(-d_{2})]$

[icon name=”bell” class=””] Determining Price of a European call option on the mixed Geometric Average $ M_t = (S_tQ_t)^{\frac{1}{2}} $ with strike K. [Payout at Expiration = $(M_t-K)_+$]

$ M_t = (S_tQ_t)^{\frac{1}{2}} $

=>  $ M_t^2 = S_tQ_t = [S_0e^{(r-\delta_s-\frac{1}{2}\sigma_s^2)t+\sigma_sZ_s \sqrt t}][Q_0e^{(r-\delta_q-\frac{1}{2}\sigma_q^2)t+\sigma_qZ_q \sqrt t}]$

=>  $ M_t = [S_0Q_0e^{(2r-\delta_s-\delta_q-\frac{1}{2}\sigma_s^2-\frac{1}{2}\sigma_q^2)t+(\sigma_sZ_s + \sigma_qZ_q )\sqrt t}]^\frac{1}{2}$

$ln M_t = ln[S_0Q_0]^\frac{1}{2} + [r-\frac{\delta_s+\delta_q+\frac{1}{2}\sigma_s^2+\frac{1}{2}\sigma_q^2}{2}]t + [\frac{\sigma_sZ_s + \sigma_qZ_q}{2}]\sqrt t$

$E[ln M_t] = ln[S_0Q_0]^\frac{1}{2} + [r-\frac{\delta_s+\delta_q+\frac{1}{2}\sigma_s^2+\frac{1}{2}\sigma_q^2}{2}]t$

$Var[ln M_t] = \sigma_m^2 = Var[[\frac{\sigma_sZ_s + \sigma_qZ_q}{2}]\sqrt t] = \frac{1}{4}[\sigma_s^2 + \sigma_q^2 +2\rho\sigma_s\sigma_q]t$

Compare $ M_t$ to the generic expression $M_0e^{(r-\delta_m-\frac{1}{2}\sigma_m^2)t+\sigma_mZ_m \sqrt t}$:

$\delta_m  + \frac{1}{2}\sigma_m^2 = \frac{[\delta_s+\delta_q+\frac{1}{2}\sigma_s^2+\frac{1}{2}\sigma_q^2]}{2}  $

Having identified $\sigma_m$ and $\delta_m$

Price of call = $c_M = [M_0e^{-\delta_m} * N(d_1) – Ke^{-\delta_q} * N(d_2)]$

Bounds for Option Prices

European Call: $F^P(S) \geq C_{Eur}(S,K,T) \geq max(0, F^P(S) – F^P(K))$

European Put: $F^P(K) \geq P_{Eur}(S,K,T) \geq max(0, F^P(K) – F^P(S))$

American Call: $S \geq C_{Amer}(S,K,T) \geq max(0, S-K)$

American Put: $K \geq P_{Amer}(S,K,T) \geq max(0, K-S)$


Call:  $S \geq C_{Amer}(S,K,T) \geq C_{Eur}(S,K,T) \geq max(0, F^P(S) – F^P(K))$

Put: $K \geq P_{Amer}(S,K,T) \geq P_{Eur}(S,K,T) \geq max(0, F^P(K) – F^P(S))$

Early Exercise of American Options

Check for American CALL

Exercising a call option entails purchasing the stock at the pre-determined ‘Strike price’. For an American call, an early exercise entails the following:

  • You no longer are able to leverage the interest on the Strike Price [K]  which you otherwise  would have gained. [Thus an effective indirect loss]
  • Forgoing the benefit of no-exercise at expiry, thus losing an implicit put option!
  • Benefiting from the stocks dividends since you now own the stock.

When is exercise optimal for an American Call ?

  • Stock does not pay dividends => Exercise is never optimal
  • Stock pays dividends
    • Early exercise may be optimal if PV(Dividends)[btw]  $= S_0 – F_{t,T}^P(S)$  [/btw] > PV(Interest on the Strike) [btw]$ = K(1-e^{-r(T-t)})$ [/btw]
    • Early exercise is optimal if PV(Dividends) > PV(Interest on the Strike) + Implicit Put

 

 

Check for American PUT

  • Irrespective of stock paying dividend
    • Early exercise may be optimal if PV(Dividends) < PV(Interest on the Strike)
    • Early exercise is optimal if PV(Dividends) < PV(Interest on the Strike) Implicit call

Market Makers' Profit

Delta Hedging

Since traders mostly purchase options, Market-makers are often exposed to short positions and hence losses are theoretically limitless.

In a Delta-Hedged portfolio, for a change from $S_t$ to $S_{t+h}$

Delta Hedging -> Short a call, Long delta shares of stock and borrow.

Market Makers’ Profit  = Change in value of $\Delta$ shares + Change in option value – Interest Expense

Market Makers’ Profit  =$ [\Delta _t (S_{t+h} – S_t) ] + [ – (C_{t+h} – C_t)] – [rh( \Delta _t S_t – C t)]$

From $\Delta \Gamma \Theta $ approximation,

$C_{t+h} – C_t  = \epsilon \Delta_t  + 0.5 \epsilon ^2 \Gamma + h \Theta$  [btw] $\epsilon  = S_{t+h} – S_t  $ [/btw]

[icon name=”bullseye” class=””] $V_{ud}-V = (S_{ud}-S) \Delta (S,0) + 0.5 (S_{ud}-S)^2 \Gamma (S,0) + 2h \Theta$

  • Since $\Gamma $ is positive for both puts and calls,  Market makers profit is decreasing  with $\epsilon ^2$ . He thus looses for huge changes in the stock price in either direction. [btw] For large changes in $ | \epsilon | $ market-makers profit is less [/btw]
  • He breaks-even when $\epsilon = \sigma \sqrt h S_t $.
    • $\epsilon < \sigma\sqrt h S_t$ – Makes Profit
    • $\epsilon > \sigma\sqrt h S_t$ – Goes into loss

Delta Gamma Hedging [Gamma neutral portfolio]

  1. Use another option on the same stock to neutralize $\Gamma $ of his existing position. [btw] Greeks in a portfolio are additive! [/btw]
  2. Calculate the $\Delta$ of the new portfolio
  3. Now Delta hedge using the underlying stock. [btw] $\Gamma$ will remain neutralized because $\Gamma _{Stock} = 0$  [/btw]

State Prices /Utility

$Q_i$ = Price of a security that pays 1\$ iff the stock moves to state $i$ = Partial expected value of PV(1)

$U_i$ = PV(1), given the stock is in state $i$

$Q_u = pU_u = \frac{p^*}{1+r}$   |   $Q_d = (1-p)U_d = \frac{1-p^*}{1+r}$ | $Q_u + Q_d = e^{-r}$

$C_0 = Q_uC_u + Q_dC_d$ | $S_0 = Q_uS_u + Q_dS_d$

$p* = \frac{Q_u}{Q_u + Q_d}$

 

 

 


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