Can anyone explain the logic behind this formula? (1+RF-D/U-D) Schweser doesn’t seem to offer any insight other than “just memorize this” Formulas seem easier to memorize though, when I at least have a little understanding of the logic behind them. Thanks!
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You can derive this formula by creating a hedged portfolio which will have a fixed value at t=1. For example, if we have a portfolio of call and delta stocks then the initial value of portfolio at t=0: C0 - (Delta)S0 At t=1, for up move value of portfolio: C+ - (Delta)S+ At t= 1, for down move value of portfolio: C- - (Delta)S- if the portfolio has same value in up move and down move then: C+ - (Delta)S+ = C- - (Delta)S- => Delta = (C+ - C-)/(S+ - S-) Since we have removed any uncertainty regarding the value of portfolio in future the present value of the portfolio should be equal to cost of setting up the portfolio. so, C0 - (Delta)S0 = (C+ - (Delta)S+)/(1 + rf) -----1 Now, Delta = (C+ - C-)/(S+ - S-) or (C+ - C-)/((u - d)*S0) and S+ = u*So relacing the above in equation - 1, we get C0 - (C+ - C-)/(u - d) = (C+ - u*(C+ - C-)/(u - d))/(1 + rf) Solve the above equation, you will get something like: C0 = (Cu*(1 + rf -d)/(u-d) + Cd*(u - 1 -rf)/(u-d))/(1 + rf) (1 + rf -d)/(u-d) = risk neutral probablity = p (u - 1 -rf)/(u-d) = 1 - p
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after having gone thro’ above technical stuff, i though it is better to remember the formula in the first post ;;phew…
kabhii derived the whole formula for you. I’m going to add a brief explanation to that. in risk-neutral world on average assets grow at risk-free rate. in example of an option, expectation of one step increment should be equal to risk-free rate (or asset that’s worth $1 becomes equal $1*(1+RFR) and if we using formule for expectation, we get: U*p + D*(1-p) = 1+RFR // p*(U-D) = 1+RFR - D p = (1+RFR-D)/(U-D)
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Maratikus, two thumbs up. Easy explanation!
Just a reminder … Use (1+r)^T - d / u-d. The period is not always 1 year.
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Thanks so much everyone for the awesome responses!
Perhaps this is a helpful addendum
Its a very useful point maratikus makes:
The current value should be the present value of the outcomes (we can ‘normalize’ using $1 as starting value):
ie [ p *(Rup) + 1-p *(Rdown) ] / (1+rfr) = current value ($1 to make the math easy)
But we want p
So you just re-write:
[p *(Rup) + 1-p *(Rdown) ] / (1+rfr) = $1
[p(Rup) + Rdown - p(Rdown) ] / (1+rfr) = $1
p(Rup) + - p(Rdown) = $1 * (1+rfr) - Rdown
p(Rup – Rdown) = $1 * (1+rfr) - Rdown
p = [(1+rfr) - Rdown ] / (Rup – Rdown)