If $W_t$ is standard Brownian motion, what is meant by $\int_0^T W_t dW_t$ in finance?

Furthermore, what then is the meaning of $\int_0^T W_t \ln(W_t) dW_t$?

  • 2
    $\begingroup$ Are you sure you mean $\int_0^T W_TdW_t$ and not $\int_0^T W_tdW_t$? Also, what do you mean with ''what do these integrals mean in finance?'' These integrals have a very clear math definition and sometimes appear in option pricing. In general, $\int \alpha_t dS_t$ is the value of a portfolio (trading strategy) with holds $\alpha_t$ many shares of $S_t$. $\endgroup$ – Alex Oct 25 '20 at 22:46
  • $\begingroup$ perhaps you can tell the other poster to correct $T$ to $t$ as well quant.stackexchange.com/questions/58923/… $\endgroup$ – develarist Oct 25 '20 at 22:52
  • $\begingroup$ Hi @develarist: It's a good question, I think intuition is always important in finance. I had a crack at the intuition behind Ito Integrals of the type $\int_0^TW_t\,dW_t$ in this answer here. Also a good answer to your question is here: MagicIsInTheChain shows that the integral of the type $\int_0^T W_TdW_t$ can be rewritten as $2\int_0^TW_t\,dW_t$, which then links back to my answer. $\endgroup$ – Jan Stuller Oct 26 '20 at 10:38
  • 6
    $\begingroup$ The reason for asking is $W_T$ can take negative values, so wanted to see the context. Have seen $\int_0^T{W_T f(W_T) dW_t}$ but not for f=ln. $\endgroup$ – Magic is in the chain Oct 28 '20 at 17:46
  • 1
    $\begingroup$ @Magicisinthechain: out of curiosity, how would one interpret the integral $\int_0^T{W_T f(W_T) dW_t}$ (or $\int_0^T{W_t f(W_t) dW_t}$) ? Where would it arise and what would it represent? $\endgroup$ – Jan Stuller Nov 8 '20 at 10:51

I am sure there will be more thorough answers provided by others, but let me have a quick go at the first part: "what is meant by $\int_0^T W_t dW_t$ in finance?".

I like to interpret Ito Integral as the outcome of a gambling strategy. In general, Ito Integral can be written as:

$$I_t:=\int_{h=0}^{h=t}f(Y_h)dX_h=\lim_{n \to\infty}\sum_{h=0}^{n-1}f(Y_h)\left(X_{h+1}-X_h\right)$$

Above, $X_t$ is a generic stochastic process (doesn't necessarily have to be $W_t$), whilst $Y_t$ is a square-integrable process (doesn't have to be stochastic). $Y_t$ has to be adapted to the filtration generated by $X_t$. $f()$ is some well-behaved function that still makes $f(Y_t)$ square integrable.

I interpret the integrator $X_t$ as the outcome of the gambling game, whilst the integrand $f(Y_t)$ is the betting strategy.

Illustrative example: let's suppose $X_h$ represents a coinflip for each $h$ (i.e. $X_h\epsilon ${$-1,1$} with probability $0.5$), $Y_h=1$ and $f()=2$. Then a discrete Stochastic integral (finite sum, strictly speaking not an Ito integral) could be defined as: $I_{t=10}=\sum_{h=0}^{9}2\left(X_{h+1}-X_h\right)$. This quantity computes the outcome of a gambling game after 10 rounds of betting, where each round the bettor bets consistently 1 unit of currency, and can either win or lose twice what he or she bets.

Moving on, taking $X_t=W_t$, $Y_t=W_t$ and $f()=1$, I interpret the Ito integral $$I_t:=\int_{h=0}^{h=t}W_hdW_h=\lim_{n \to\infty}\sum_{h=0}^{n-1}W_h\left(W_{h+1}-W_h\right)$$

as the outcome of a betting game, where initially the bettor bets $W_0:=0$, but each subsequent moment in time, the bettor bets the realized sum (up to that point in time) of Brownian increments $W_{h+1}-W_h$. These Brownian increments are at the same time the gambling game pay-off (so the game pays the bettor's bet multiplied by the next Brownian increment realization).

In continuous time, the bettor constantly adjusts his or her bet to the "current" level of the Brownian motion $W_t$, which acts as the integrator: i.e. the betting game pays the realized Brownian $W_t$ at each moment in time multiplied by the bettor's bet corresponding to the last observed realization of $W_t$.

Finally, if the integrator is some stock price process $S_t$ instead of $W_t$, and $f(Y_t)$ is the number of stocks held (could be simply a constant, deterministic quantity), then I interpret the Ito Integral as the profit or loss of that stock portfolio over time.

  • $\begingroup$ Any take on the logarithmic part? $\endgroup$ – develarist Nov 8 '20 at 8:23
  • 1
    $\begingroup$ @develarist: as per some of the comments above, log of a negative number is undefined. Brownian motion can take on negative values, so log of a Brownian motion is not a well defined quantity, I am not sure how to interpret it. $\endgroup$ – Jan Stuller Nov 8 '20 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.