I am really having a terrible time applying Girsanov's theorem to go from the real-world measure $P$ to the risk-neutral measure $Q$. I want to determine the payoff of a derivative based an asset which is paying dividends, so the dividends will affect the value of the asset by changing the drift negatively, but aren't considered when calculating the payoff because it's just derived from the asset prices over time.

We begin in the real-world measure $P$. We are interested in some index, $S_t$, which I believe we can just treat as a single entity like a stock and which we can observe in the real world. It has volatility $\sigma$. We also know that this index is paying dividends continuously at a rate of $\delta$. The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on, so I take $\mu$ to be $0 - \delta$. First question: Is it normal to assume no other drift?

$dSt = (-\delta) St dt + \sigma St dWt$

Where $dWt$ are standard Wiener increments.

So clearly under $P$, our index asset has drift $-\delta$ because it is paying dividends out.

If someone actually holds the asset then they can reinvest those dividends, but we are only concerned with the value of the asset in order to determine the payoff of a derivative.

So what we want to do now is determine the expected value of the asset under the risk neutral measure, $Q$. We want this expected value, discounted by the value of a bond earning the risk free rate, to be a martingale in this measure, so the asset must also drift at the risk free rate. In this way, an asset under the risk neutral measure with volatility 0 is equivalent to a bond.

In order to achieve this, we want to change the drift of our process from $-\delta$ under $P$ to $r$ under $Q$. We define a new process

$X_t = \theta t + W_t$


$\theta = \frac{(-\delta - r)}{\sigma}$

If we now consider the discounted asset price $S_t^* = S_t / Bt = S_t e^{-rt}$ we have the following.

$dSt^* = (-\delta - r) St^* dt + \sigma St^* dWt$

$dSt^* = (-\delta - r) St^* dt + \sigma St^* (X_t - \theta)$

$dSt^* = (-\delta - r) St^* dt + \sigma St^* (X_t - \frac{-\delta - r}{\sigma})$

$dSt^* = \sigma St^* X_t$

Where now we have the process $St^*$ as a geometric Brownian with no drift if we are under $Q$ and with the same drift as before under $P$. All we have done was put $(X_t-\theta)$ in for $W_t$ and these are equal by definition.

So under $Q$, and without discounting, the process for the asset $S_t$ should follow

$dSt = r St dt + \sigma St dXt$

The dividend being payed by the asset is thus somehow 'inside' the brownian motion $X_t$ and that's apparently all we see of it?

At this point, I'd like to stop and ask "at what point do we consider ourselves 'under $Q$'?". By google-fu, I found an example program in R where they make a binomial approximation to Brownian motion by taking constant size steps up or down with different propbabilities. The relevant portion of code looks like:

n = 2000
t = (0:n)/n  # [0/n, 1/n, .... n/n]
dt = 1/n
theta = 1
p = 0.5 * (1 - sqrt(dt) * theta)

u = runif(n)  # Random uniform variates
dWP = ((u < .5) - (u > .5))*sqrt(dt)    # increments under P
dWQ = ((u < p) - (u > p))*sqrt(dt)      # increments under Q

WP[1:n+1] = cumsum(dWP)
WQ[1:n+1] = cumsum(dWQ)
XP = WP + theta*t      # Theta exactly offsets the change in measure
XQ = WQ + theta*t      # Now XQ == WP

I see that they are using $p$ instead of 0.5 to make decisions about the sign in the binary discretization, but if we have access to actual random normal variates is this necessary? Shouldn't it be a enough to simply define $\theta$ appropriately as above and then use it and $W_t$ to derive $X_t$, directly? Although, if we do that, then we go from $W$ to $X$ but I don't understand how we got from $P$, to $Q$ if that happened at all.

The payoff of our asset then, is the discounted expectation under the risk-neutral measure. So we evolve the process $S_t$ under measure $Q$ until expiry time T, calculate the payoff. Repeat this a few thousand times and then take the average and discount it back to $t = 0$ to get the price.

I thought all of this seemed pretty reasonable and was making sense, but when I try to implement it, I get nothing like what I expect. Can anyone confirm that the story I just presented about how we arrive at the risk neutral measure is correct?

A more general question: $W_t$ is a martingale under $P$. We shift it and change measures to create $X_t$ and have that be a martingale under $Q$. We then model the asset based on the process $X_t$ with drift r. If all we wanted was a process under which the stock drifts at the risk free rate, what do we gain by changing from $W$ to $X$ and $P$ to $Q$ if they exactly offset each other? Why don't we just use $W_t$? And what happened to the dividend?

If the stock process is drifting downward at $-\delta dt$ under $P$, what does it really look like under $Q$? It can't just look like it's drifting at the risk-free-rate because then the dividend would have had no effect at all.

Further search produced some suggestions that it should be drifting at $r-\delta$, because we assume that the dividends are invested into bonds. Why would we make such an assumption? Is it because we want to consider a portfolio containing the asset and not the asset itself?


ad) "Is it normal to assume no other drift?" Under measure P you might have drift. You could use it as a working assumption, but in general indices drift every now and then. So, no, usually you do not assume away the drift.

"The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on" geometric Brownian motion: S follows a gBm if it satisfies the well-known dS/S SDE, drift non-zero, contrary to your deduction.

ad) "at what point do we consider ourselves 'under Q '?". Q is the risk-neutral measure: when you insert your X-expression for W, you switch to drift r and to Q. Btw, see the link for getting notation and derivation right.

ad) "If the stock process is drifting downward at −δdt under P , what does it really look like under Q ?" Insert mu=mu*-delta in here. This yields drift coefficient (r+delta)S_t under X.

ad) "because we assume that the dividends are invested into bonds." Here the dividend is an average yield over the holding period. You reinvest it in the underlying.

ad) "I thought all of this seemed pretty reasonable and was making sense": Sounds good.

  • $\begingroup$ So if I am understanding you correctly here, the dividend payments will result in the stock drifting higher under Q than it would without them: (r+delta) instead of (r-delta)? I don't see how that works. Taking the dividends out of the stock should lower the value of the stock right? If they are being reinvested into the stock, our portfolio will reflect that we own more shares, but I don't see why the value of the stock should also be going up. $\endgroup$
    – John Tyree
    May 18 '12 at 13:40
  • $\begingroup$ Also, when you say that "S follows a gBm if it satisfies the well-known dS/S SDE, drift non-zero...", I assume you are referring to dS/S = mu dt + sig dz, but why does this stop being gBm if mu = 0? $\endgroup$
    – John Tyree
    May 18 '12 at 13:43
  • $\begingroup$ ad) "I assume you are referring to dS/S = mu dt + sig dz, but why does this stop being gBm if mu = 0?": In "The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on" you seemed to deduce that mu=0 from the fact that the index is supposed to follow a gBm. I wanted to make clear to you, that although the index is supposed to follow a gBm, this does not imply that mu=0. $\endgroup$
    – Konsta
    May 18 '12 at 18:28
  • $\begingroup$ Yes, you are right. It should've been mu=mu*-r, i.e. theta=(mu*-r)/sigma with mu*=0 with your gBm assumption. This would've yielded (r-delta) under X. $\endgroup$
    – Konsta
    May 18 '12 at 21:34
  • $\begingroup$ Ok I think I understand now. The drift is not 0, but it will be incorporated into the process X under Q, such that the SDE will always take the form that you described with (r-delta). $\endgroup$
    – John Tyree
    May 18 '12 at 23:01

Just following Musiela Rutkowski (the link redirects to Amazon). The risk neutral measure is derived form imposing that the present value of a self financed portfolio (i.e.; no infusion or withdraw of money) is a martingale. A portfolio can be seen as a stochastic process where its value at time $t$ is given by $$ V_t = \phi^0_tP_t + \phi^1_tS_t\ , $$ this is, the value is equal to the number of bonds $\phi^0_t$ times the growth of the bond $P_t=e^{rt}$ plus the number of shares $\phi_t^1$ times the value of the share $S_t$ at time $t$. If we assume a constant dividend yield $q$ (proportional to the spot) the increment in value is given by $$ dV_t = \phi_t^0 dP_t + \phi_t^1dS_t + \phi_t^1qS_tdt $$ that is the change of value of every component plus the dividend amount we receive $\phi_t^1qS_t$ at every instant increment $dt$. We are assuming a geometric Brownian motion for the price of the share described by $dS_t=S_t(\mu dt + \sigma dW_t)$ under the physical measure, therefore $$ dV_t = \phi_t^0 dP_t + \phi_t^1S_t((\mu + q)dt + \sigma dW_t)\ . $$ Now just follow the same steps as the non dividend paying stock. To make the discounted value of our portfolio follow a martingale we will need $$ d(e^{-rt}V_t) = \phi_t^1dS^*_t\ ,\quad \mbox{with},\qquad dS_t^*=S_t^*\sigma dX_t $$ $$ S_t^*=\frac{1}{P_t}e^{qt}S_t $$ $$ X_t = W_t - \frac{r-q-\mu}{\sigma}t $$ Replacing $dW_t=dX_t + \frac{1}{\sigma}(r-q-\mu)dt$ in the equation of $dS_t$ yields $$ dS_t = S_t((r-q)dt + \sigma dX_t) $$ the evolution under the risk neutral measure.


I love Bjork's explanation on this in his book "Arbitrage theory in continuous time". Although it's best to read his own words, I'll try to summarize the gist of it below:

  1. First of all, only tradable assets have a risk-neutral expected return equal to the risk-free rate
  2. The price process of a dividend paying stock is not tradable, meaning if you just hold the stock $S$, unless you give away the dividends for free, which would make you an irrational investor (and arbitrage pricing assumes rational investors otherwise there would be immediate arbitrage), your gains process is not the price gains of $S$ only, but includes the accrued dividends. So the tradable quantity here is not $S$ but $Se^{qt}$. In other words the risk-neutral drift of $Se^{qt}$ is $r$. In other words, under the risk-neutral measure

$$ d \left(Se^{qt} \right) = r Se^{qt} dt + \sigma Se^{qt} dW $$

or equivalently

$$ dS = (r-q)Sdt + \sigma S dW $$


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