2

Given

\begin{align*} dX_{t} &= \mu dt + \sigma X_{t}dB_{t}\\ \log(X_{t}) &={-\frac{1}{2}\int_{0}^{t} \sigma^{2}ds}+{\int_{0}^{t} \sigma dB_{s}} \end{align*}

I am trying to set $\log(Y_{t}) := \frac{1}{2}\sigma^{2}t - \sigma B_{t}$ and show that

if $Z_{t} = X_{t}Y_{t}$, then $dZ_{t} = \mu Y_{t} dt$.

However, I am not very confident in how to calculate $dY_{t}$.

I defined $f(x,y) = xy$ and wrote

$$df = \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial x}\frac{\partial f}{\partial y}dxdy = xdy + ydx + dxdy$$

so that under Ito,

$$dZ_{t} = X_{t}dY_{t} + Y_{t}dX_{t} + d{X_{t}}d{Y_{t}}.$$

The term $Y_{t}dX_{t}$ is straightforward:

$$Y_{t}dX_{t} = \mu Y_{t}dt + \sigma X_{t}Y_{t}dB_{t} = \mu Y_{t}dt + \sigma dB_{t}.$$

However, I have no idea how to approach computing $dY_{t}$. I have tried to write the following which I do not think is correct:

$$d\log(Y_{t}) = \frac{dY_{t}}{Y_{t}} = \frac{1}{2}\sigma^{2}dt -\sigma dB_{t}$$

so that

$$dY_{t} = \frac{1}{2}\sigma^{2}Y_{t}dt - \sigma Y_{t}dB_{t}.$$

Then substituting this all in:

$$X_{t}dY_{t} = \frac{\sigma^{2}}{2}X_{t}Y_{t}dt - \sigma X_{t}Y_{T}dB_{t} = \frac{\sigma^{2}}{2}dt - \sigma dB_{t}$$

and finally

$$dX_{t}dY_{t} = \left(\mu dt + \sigma X_{t}dB_{t}\right) \left(\frac{1}{2}\sigma^{2}Y_{t}dt - \sigma Y_{t}dB_{t}\right) = -\sigma^{2}X_{t}Y_{T}\langle B_{t},B_{t}\rangle = -\sigma^{2}dt$$

which gives

$$dZ_{t} = \frac{\sigma^{2}}{2}dt - \sigma dB_{t} + \mu Y_{t}dt + \sigma dB_{t} -\sigma^{2}dt = -\frac{\sigma^{2}}{2} + \mu Y_{t}dt.$$

So I am probably off by a factor of $2$ somewhere and I suspect it is in the computation of $dY_{t}$, but I am not sure. What am I doing wrong?

JessicaK
  • 7,655

1 Answers1

3

It seems to me that you are using several times $X_t Y_t = 1$ (for instance when you compute $Y_t \, dX_t$ you are claiming that $\sigma X_t Y_t \, dB_t = \sigma \, dB_t$). This is wrong; please check your calculations.

Regarding the factor $1/2$: You didn't calculate $dY_t$ correctly. Note that Itô's formula states that

$$df(Y_t) = f'(Y_t) \, dY_t + \frac{1}{2} f''(Y_t) \, (dY_t)^2$$

and therefore

$$d\log(Y_t) \neq \frac{dY_t}{Y_t}.$$

If you apply Itô's formula correctly, you will find that

$$d\log(Y_t) = \frac{dY_t}{Y_t} - \frac{1}{2} \frac{(dY_t)^2}{Y_t^2}. \tag{1}$$

If we assume for the moment being that $$dY_t = f(t) Y_t \, dt + g(t) Y_t \, dB_t \tag{2} $$ for suitable mappings $f,g$, then $(dY_t)^2 = Y_t^2 g(t)^2 \, dt$, and $(1)$ gives

$$\frac{1}{2} \sigma^2 dt - \sigma dB_t = d\log(Y_t) = \left( f(t)-\frac{1}{2} g(t)^2 \right) \, dt + g(t) \, dB_t.$$

Thus,

$$g(t) = - \sigma \quad \text{and} \quad f(t) = \frac{1}{2} \sigma^2 + \frac{1}{2} g(t)^2 = \sigma^2$$

implying, by $(2)$,

$$dY_t = \sigma^2 Y_t \, dt - \sigma Y_t \, dB_t.$$

Alternative approach to calculate $dY_t$: By the very definition of $\log(Y_t)$, we have

$$Y_t = \exp(\log(Y_t)) = \exp \left( \frac{1}{2} \sigma^2 t - \sigma B_t \right).$$

Applying Itô's formula for $g(t,x) := \exp(\sigma^2 t/2- \sigma x)$, we find

$$dY_t = dg(t,B_t) = - \sigma g(t,B_t) \, dB_t + \left( \frac{\sigma^2}{2} g(t,B_t) + \frac{\sigma^2}{2} g(t,B_t) \right) \, dt = - \sigma Y_t \, dB_t + \sigma^2 Y_t \, dt.$$

saz
  • 120,083
  • If $Y_{t} = \operatorname{exp}(\frac{1}{2}\sigma^{2}t - \sigma B_{t})$ and $X_{t} = \operatorname{exp}(-\frac{1}{2}\sigma^{2}t + \sigma B_{t})$, I have a hard time seeing why it is the case that $X_{t}Y_{t} \neq exp((\frac{1}{2}\sigma^{2}t - \sigma B_{t} + ( -\frac{1}{2}\sigma^{2}t + \sigma B_t))) = e^{0} = 1$. Now that you mention it, it clearly cannot be true because if it were, $Z_t = X_tY_t = 1$ and $dZ_t$ would be trivial. I can only guess that $-\sigma B_t + \sigma B_t\neq 0$. Is it due to the reflection principle that this becomes $2\sigma B_{t}$? So that $X_tY_t = e^{2\sigma B_t}$? – JessicaK Jan 15 '18 at 23:53
  • @JessicaK Why do you think that $X_t= \exp(-\sigma^2 t/2 + \sigma B_t)$...? That's not true; note that there is a constant $\mu$ in the very definition of $X_t$. – saz Jan 16 '18 at 08:30
  • Ah, everything is clear now. Thank you very much for the help. – JessicaK Jan 16 '18 at 15:49
  • @JessicaK You are welcome. – saz Jan 16 '18 at 15:54