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Since options represent leveraged stock investments, at which strike $K$ does a European option provide maximum leverage?

Hereby define leverage $L$ as ratio of Delta/Optionprice:

$$L(K)=\frac{\Delta(K)}{C(K)}$$

You can assume all parameters fixed and positive ($T-t>0$) except strike $K (>0)$.

Delta is defined as $\Delta=\frac{\partial C(S)}{\partial S}$.

The maximum option leverage strike is important as it provides the maximum possible profit (and loss) on investment.


  • Numerical solutions would be acceptable (e.g. MATLAB fmincon).

  • Graphical solutions would also be acceptable (e.g. MATLAB plot or http://www.wolframalpha.com/input/?i=x%5E2).

  • Intuitive explanations would be acceptable.

  • For theoretical solutions you can use Black-Scholes model where

\begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - t)\right] \\ &= d_1 - \sigma\sqrt{T - t} \end{align} and \begin{align} P(S, t) &= Ke^{-r(T - t)} - S + C(S, t) \\ &= N(-d_2) Ke^{-r(T - t)} - N(-d_1) S \end{align}

The deltas for call and put are

$$\Delta^C=N(d_1)$$ $$\Delta^P =N(d_1) - 1$$

where $N(\cdot)$ denotes the cumulative Normal distribution.

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  • 2
    $\begingroup$ The classic definition of leverage for options is a little bit different. It is 'lambda' defined as Delta times (Stockprice/Optionprice). See for example ericbenhamou.net/documents/Encyclo/… . It makes sense to include Stockprice I think. $\endgroup$ – Alex C Sep 27 '15 at 1:37
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    $\begingroup$ @AlexC Ok please additionally assume a fixed current stockprice $S_t$, however then it doesnt matter whether you multiply with $S_t$ or not. $\endgroup$ – emcor Sep 27 '15 at 10:12
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    $\begingroup$ So you're basically looking for an analytical solution for $\frac{\partial \frac{\Delta}{C(S,t)}}{\partial K} = 0$, right? $\endgroup$ – SRKX Sep 29 '15 at 12:19
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    $\begingroup$ @SRKX But one can only use the first order condition if the leverage function is convex in $K$. $\endgroup$ – emcor Sep 29 '15 at 12:47
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Bob Jansen Oct 13 '15 at 15:25
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Ciao, I'm studying this problem from a while. Let me post the graph obtained numerically. I've used the following parameters:

$$ \left\{ \begin{array}{rcl} S &=& 2 \\ r &=& 0.01 \\ \sigma &=& 0.2 \\ K &\in& [0.1, 10] \\ T &=& 5 \\ t &=& 1 \end{array} \right. $$

enter image description here

This is not good since the function is always increasing and that means that it has no max value wrt $K$.

The Put case is really intersting, let me report the plot: enter image description here

Of course in this case we have a minimum (working very hard for a closed form...in coming). I solved the minimum problem numerically again (in the Put case). In order to do it I've studied the behaviour of the minimum value of $K$ wrt $S$. In turns out that the following equation hold: $$ K_{min} = \frac{3}{2}S $$ enter image description here

This is the R code I've used:

d1 <- function(S, K, sigma, r, T, t){
  return(1/(sigma*sqrt(T-t))*(log(S/K) + (r + sigma^2/2)*(T-t)))
}

d2 <- function(S, K, sigma,r,  T, t){
  return(d1(S, K, sigma, r, T, t) - sigma*sqrt(T-t) )
}

Call <- function(S, K, sigma,r,  T, t){
  return(pnorm(d1(S, K, sigma, r, T, t))*S - pnorm(d2(S, K, sigma, r, T, t))*K*exp(-(T-t)) )
}

DeltaCall <- function(S, K, sigma, r, T, t){
  return(pnorm(d1(S, K, sigma, r, T, t)))
}


Put <- function(S, K, sigma, r, T, t){
  return(K*exp(-r*(T-t)) - S + Call(S, K, sigma, r, T, t))
}

DeltaPut <- function(S, K, sigma, r, T, t){
  return(DeltaCall(S, K, sigma, r, T, t) - 1)
}

leverageCall <- function(S, K, sigma, r, T, t){
  return(DeltaCall(S, K, sigma, r, T, t)/Call(S, K, sigma, r, T, t))
}

leveragePut <- function(S, K, sigma, r, T, t){
  return(DeltaPut(S, K, sigma, r, T, t)/Put(S, K, sigma, r, T, t))
}


minLPut <- function(S, K, sigma, r, T, t){
  LPut = leveragePut(S, K, sigma, r, T, t)
  return(min(LPut))
}

argminLput <- function(S, K, sigma, r, T, t){
  LPut = leveragePut(S, K, sigma, r, T, t)
  return(K[LPut == min(LPut)])
}



S = 2
r = 0.01
sigma = .2
T  = 5
t = 1
K = seq(0.1, 10, by = 0.01)

LCall = leverageCall(S, K, sigma, r, T, t)
LPut = leveragePut(S, K, sigma, r, T, t)

argminLput(S, K, sigma, r, T, t)
argminLput(4, K, sigma, r, T, t)

plot(K, LCall,
     xlab = "Strike",
     ylab = "Leverage",
     type = "l",
     main = "Call Leverage")
grid()

plot(K, LPut,
     xlab = "Strike",
     ylab = "Leverage",
     type = "l",
     main = "Put Leverage")
grid()



S = seq(0.1, 10, by = 0.01)
yS = S
for(i in 1:length(S)){
  yS[i] = argminLput(S[i], K, sigma, r, T, t)
}

plot(S, yS,
     type = "l",
     xlab = "Spot",
     ylab = "argmin K",
     main = "Put argmin plot")
grid()
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  • 2
    $\begingroup$ Thanks, its highly interesting. It appears that the "optimal" point is at "K=S" where you get the most marginal leverage increase. So ATM options are somewhwat optimal in this sense. It also appears logical that the more risk you take with OTM options, the higher is your implied leverage. Could you maybe post the same graph for the put option also? Thanks, $\endgroup$ – emcor Dec 21 '17 at 12:41

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