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Given stochastic processes $X_t, Y_t$ of form given below, we want to find $d(X_t Y_t)$. Assume $$X_t=\sigma_tW_t + \mu_t t$$ $$Y_t=\rho_tW_t + \nu_t t$$

I have two different answers depending on how I derive them, and both of them are wrong, but I don't know why.

Method 1:

$X_tY_t=\sigma_t\rho_tW_t^2+(\sigma_t\nu_t+\mu_t\rho_t)tW_t+\mu_t\nu_tt^2$

Therefore $X_tY_t=f(W_t,t).$ Using Taylor expansion: $df(W_t,t)=\frac{\delta f}{\delta W_t}dW_t+(\frac{1}{2}\frac{\delta^2f}{\delta W_t^2}+\frac{\delta f}{\delta t})dt.$

Therefore: $d(X_tY_t)=(2\sigma_t\rho_tW_t+\sigma_t\nu_t+\mu_t\rho_t)dW_t+2(\sigma_t\rho_t+\mu_t\nu_tt)dt$

Method 2:

However, if we use a different approach, by defining instead $g(X_t,Y_t)=X_tY_t$, then a Taylor polynomial gives: $$d(X_t,Y_t)=\frac{\delta g}{\delta X_t}dX_t+\frac{\delta ^2g}{\delta X_t^2}(dX_t)^2+\frac{\delta g}{\delta Y_t}dY_t+\frac{\delta^2 g}{\delta Y_t^2}(dY_t)^2+\frac{\delta^2 g}{\delta X_t\delta Y_t}dX_tdY_t$$ This results in: $$d(X_tY_t)=Y_tdX_t+X_tdY_t+2dX_tdY_t$$ If I were to expand this, it would be quite big and definitely not equal to the result from (1).

Both of these are wrong, because my textbook gives an even different result, namely $$d(X_tY_t)=Y_tdX_t+X_tdY_t+ \sigma_t\rho_tdt$$ I don't know how exactly they got there other than that they used $X_tY_t=\frac{1}{2}((X_t+Y_t)^2 - X_t^2 - Y_t^2)$.


I must be fundamentally misunderstanding something about stochastic calculus (I've only just started with it).

What am I doing wrong, and what misunderstanding do I have?

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  • $\begingroup$ In your method 2, you missed a factor of $\dfrac{1}{2}$ for the $(dX_t)^2$ and $(dY_t)^2$ term. $\endgroup$ Commented Mar 29, 2017 at 17:15
  • $\begingroup$ Are you sure that $X_t=\sigma_tW_t+\mu_t t$ is correct? The equation $dX_t=\sigma_t dW_t+\mu_t dt$ seems more reasonable. $\endgroup$
    – Handschuh
    Commented Mar 29, 2017 at 17:29
  • $\begingroup$ @ChrisVarghese, pff.. ok that explains it. $\endgroup$
    – user56834
    Commented Mar 29, 2017 at 17:39
  • $\begingroup$ @Handschuh, Isn't the second equation derivable from the first? It is according to my textbook, if I'm not misinterpreting it. $\endgroup$
    – user56834
    Commented Mar 29, 2017 at 17:39
  • $\begingroup$ The equation $dX_t=\sigma_t dW_t+\mu_t dt$ is shorthand for an integral equation $X_t-X_0=\int_0^t \sigma_s dW_s+\int_0^t \mu_s ds$ (this is written "everywhere", e.g. en.wikipedia.org/wiki/…). So, for example $dX_t=W_tdW_t$ means $X_t=X_0+\frac{1}{2}W_t^2-\frac{t}{2}$. $\endgroup$
    – Handschuh
    Commented Mar 29, 2017 at 20:02

1 Answer 1

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Let's begin from the end. You have the stochastic product rule $$ d(X_t Y_t) = X_t dY_t + Y_t dX_y + dX_t dY_t, $$ which is identical to your text book result if you note that $dX_t = \sigma_t$ and $dY_t = \rho_t$.

As for your method 1, you took the derivative wrong... To illustrate, $$ d\left(\sigma_t \rho_t W_t^2 \right) = 2 \sigma_t \rho_t W_t dW_t + \sigma_t \rho_t dt $$ while $$ d\left[\left(\sigma_t \nu_t + \mu_t \rho_t\right)tW_t\right] = \left(\sigma_t \nu_t + \mu_t \rho_t\right)tdW_t + W_t \frac{d\left[(\sigma_t \nu_t + \mu_t \rho_t)t\right]}{dt}dt $$

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