[Math] Application of Ito’s formula to log and exponential

stochastic-analysisstochastic-calculus

Let $X$ be a strictly positive continuous semimartingale with $X_0 = 1$ and
define the process $Y$ by
$$ Y_t = \int_0^t \frac{1}{X} dX – \frac12 \int_0^t \frac{1}{X^2} d \langle X \rangle. $$ Let the process $Z$ be given by $Z_t = e^{Y_t}$.

I now want to compute $d Z_t$ and show that $Z=X$. To compute $dZ_t$, I noticed that $Y$ follows by defining $Y_t = f(X_t) = \log(X_t)$ then by Ito's formula this gives us precisely that
$$ dY_t = d\log(X_t) = f'(X_t) d X_t + \frac12 f''(X_t)d \langle X_t \rangle = \frac{1}{X_t} d X_t – \frac12 \frac{1}{X_t^2}d \langle X_t \rangle,$$
which is the same as above but in differential notation. Then to compute $dZ_t$ I defined $g(Y_t) := e^{Y_t} = Z_t$ such that
\begin{align*} d Z_t &= d \exp(Y_t) = g'(Y_t)d Y_t + \frac12 g''(Y_t) d \langle Y \rangle_t \\
&= g'(f(X_t))d Y_t + \frac12 g''(f(X_t)) d \langle Y \rangle_t \\
&= f'(X_t)g(f(X_t))d Y_t + \frac12 (f'(X_t))^2g(f(X_t)) d \langle Y \rangle_t \\
&= dY_t + \frac12 \frac{1}{X_t} d \langle Y \rangle_t.
\end{align*}
I now wonder is this a correct derivation and how could I prove that $Z = X$? I guess I could prove that $dZ_t = dX_t$ but writing out $dY_t$ does not help me to obtain this. Thanks for any help.

Best Answer

As @Did points out, by definition of $Y_t$ in terms of stochastic integrals,

$$ \text dY_t = \frac{1}{X_t}\text dX_t-\frac{1}{2}\frac{1}{X_t^2}\text d\langle X\rangle_t\tag{1} $$ And $$ Z_t^{-1}\text dZ_t = Z_t^{-1}\text de^{Y_t}=Z_t^{-1}e^{Y_t}(\text dY_t+\frac{1}{2}\text d\langle Y\rangle_t) = \text dY_t+\frac{1}{2}\text d\langle Y\rangle_t\tag{2} $$

Using (1) in (2), conclude.

Related Solutions

Stochastic Differential Equations – Solution to General Linear SDE

Here is the complete solution to the problem including some special cases for an easy start. With analogy to the integrating factor method from ODEs it seems natural to rearrange \begin{align*} \mathrm{d}X_t = (a(t)X_t+ b(t)) \mathrm{d}t + (g(t)X_t+ h(t))\mathrm{d}B_t \end{align*} to the form \begin{align*}\mathrm{d}X_t - X_t \left( a(t)\mathrm{d}t+g(t)\mathrm{d}B_t\right) = b(t) \mathrm{d}t + h(t)\mathrm{d}B_t .\end{align*} Now we want to find a "nice" stochastic process $Z_t$ such that \begin{align*}( \star) \ \ \ \mathrm{d}(X_tZ_t) =& Z_t\mathrm{d}X_t \underbrace{- Z_tX_t \left( a(t)\mathrm{d}t+g(t)\mathrm{d}B_t\right)}_{X_t\mathrm{d}Z_t + \mathrm{d}X_t\mathrm{d}Z_t} \\ =& Z_t(b(t)\mathrm{d}t + h(t)\mathrm{d}B_t).\end{align*} Assume that $Z_t$ is an Itô process such that $$\mathrm{d}Z_t = f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t , \ \ Z_0 = 1.$$ Let us apply Itô's product formula to $\mathrm{d}(X_tZ_t)$ we obtain that \begin{align*}\mathrm{d}(X_tZ_t) =& Z_t \mathrm{d}X_t + X_t \mathrm{d}Z_t + \mathrm{d}X_t \mathrm{d}Z_t \\ ( \star \star) \ \ \ \ \qquad = & Z_t\mathrm{d}X_t + X_t \left( f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right) + (g(t)X_t + h(t))f_2(t, Z_t) \mathrm{d}t.\end{align*} Comparing the above with the right hand-side of $(\star)$ we arrive at \begin{align*}- Z_tX_t \left( a(t)\mathrm{d}t+g(t)\mathrm{d}B_t\right) = X_t \left( f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right) + (g(t)X_t + h(t))f_2(t, Z_t) \mathrm{d}t \end{align*} and thus \begin{align*} -Z_tX_tg(t)\mathrm{d}B_t &= X_t f_2(t, Z_t)\mathrm{d}B_t \\ -Z_tX_ta(t)\mathrm{d}t &= \big(X_tf_1(t, Z_t)+(X(t)g(t)+h(t))f_2(t,Z_t)\big)\mathrm{d}t.\end{align*} From the first equation we can deduce that $f_2(t, Z_t) = -Z_tg(t)$ and so the second one converts to $$-Z_tX_ta(t)\mathrm{d}t = \big(X_tf_1(t, Z_t)-Z_tg(t)(X(t)g(t)+h(t))\big)\mathrm{d}t,$$ and so $$ Z_t (-X_ta(t)+X(t)g^2(t)+g(t)h(t)) \mathrm{d}t = X_tf_1(t, Z_t)\mathrm{d}t.$$ Hence, $$f_1(t, Z_t) = Z_t(-a(t)+g^2(t)+X_t^{-1}g(t)h(t)).$$

However, we want $Z_t$ to be free of $X_t$ for that we consider two cases

CASE 1. $g(t) \neq 0$, $h(t)=0$

Then $Z_t$ would satisfy $$\mathrm{d}Z_t = Z_t(-a(t)+g^2(t))\mathrm{d}t -Z_t g(t)\mathrm{d}B_t , \ \ Z_0 = 1$$ (We solve the above SDE by a standard trick involving Ito's formula, that is, first we divide by $Z_t$ to obtain the expression for $Z_t^{-1}\mathrm{d}Z_t$ and then derive the expression $\mathrm{d}(\ln(Z_t))$)

and so $$Z_t = \exp\left( \int_0^t\left( \frac{1}{2}g^2(s) - a(s)\right) \mathrm{d}s - \int_0^t g(s)\mathrm{d}B_s\right).$$ This is the integrating factor which you obtained, and it is not the correct one if $h(t) \neq 0$.

If we continue then we obtain \begin{align*}\mathrm{d}(Z_tX_t) =& Z_tb(t)\mathrm{d}t \\ Z_tX_t =& X_0 + \int_0^t Z_sb(s) \mathrm{d}s\\ X_t =& X_0Z_t^{-1}+ Z_t^{-1}\int_0^t Z_sb(s) \mathrm{d}s\\ X_t =& \exp\left( \int_0^t\left( a(s) - \frac{1}{2}g^2(s) \right) \mathrm{d}s + \int_0^t g(s)\mathrm{d}B_s\right) \\ &\cdot \left(X_0+ \int_0^{t} b(s)\exp\left( \int_0^s\left( \frac{1}{2}g^2(r) - a(r)\right) \mathrm{d}r - \int_0^s g(r)\mathrm{d}B_r\right)\mathrm{d}s\right). \end{align*}

CASE 2. $g(t) = 0$, $h(t) \neq 0$

Then $(Z_t)$ satisfies $$\mathrm{d}Z_t = -a(t)Z_t\mathrm{d}t, \ \quad Z_0 =1 $$ and so $$Z_t = \exp\left( -\int_0^t a(s)\mathrm{d}s\right).$$

Therefore, \begin{align*}\mathrm{d}(X_tZ_t) =& Z_tb(t)\mathrm{d}t+ Z_t h(t)\mathrm{d}B_t \\ X_tZ_t =& X_0 + \int_0^t Z_sb(s) \mathrm{d}s + \int_0^t Z_sb(s) \mathrm{d}B_s\\ X_t = & X_0Z_t^{-1}+ Z_t^{-1}\left( \int_0^t Z_sb(s) \mathrm{d}s + \int_0^t Z_sb(s) \mathrm{d}B_s\right) \\ X_t = & X_0 e^{ \int_0^t a(s)\mathrm{d}s}\\ &+ e^{ \int_0^t a(s)\mathrm{d}s}\left( \int_0^t e^{ -\int_0^s a(r)\mathrm{d}r}b(s) \mathrm{d}s + \int_0^t e^{ -\int_0^s a(r)\mathrm{d}r}b(s) \mathrm{d}B_s\right). \end{align*}

The general case

So far we were not able to find the solution for the general case, to find it we need to modify and rearrange our initial equation in a slightly different way. Let write the SDE for $(X_t)$ as follows \begin{align*}\mathrm{d}X_t = (a(t)X_t + b(t)+g(t)h(t) - g(t)h(t))\mathrm{d}t + (g(t)X_t+h(t))\mathrm{d}B_t \end{align*} after rearranging we have \begin{align*}\mathrm{d}X_t - \left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)= (b(t)- g(t)h(t))\mathrm{d}t + h(t)\mathrm{d}B_t \end{align*} Now let ($Z_t$) be an Ito process that satisfies $$\mathrm{d}Z_t = f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t , \ \ Z_0 = 1.$$ After multiplying our SDE by $Z_t$ we obtain \begin{align*}Z_t\mathrm{d}X_t - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)= Z_t \left((b(t)- g(t)h(t))\mathrm{d}t + h(t)\mathrm{d}B_t\right) \end{align*} Applying Ito product formula to $X_tZ_t$ we get $$ \mathrm{d}(X_tZ_t) = Z_t\mathrm{d}X_t + X_t\mathrm{d}Z_t+ \mathrm{d}X_t\mathrm{d}Z_t,$$

we want to have \begin{align*}X_t\mathrm{d}Z_t+ \mathrm{d}X_t\mathrm{d}Z_t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*}

The LHS equals to \begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right)+ (g(t)X_t+h(t))f_2(t, Z_t)\mathrm{d}t.\end{align*}

We compare with the RHS \begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right)+ (g(t)X_t+h(t))f_2(t, Z_t)\mathrm{d}t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*}

and conclude that $f_2(t, Z_t) = -g(t)Z_t$. Now

\begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t -g(t)Z_t\mathrm{d}B_t \right)- (g(t)X_t+h(t))g(t)Z_t\mathrm{d}t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*} simplifies to \begin{align*}X_tf_1(t, Z_t)\mathrm{d}t -g(t)^2Z_tX_t\mathrm{d}t = - a(t)Z_tX_t\mathrm{d}t\end{align*} and so \begin{align*}X_tf_1(t, Z_t)\mathrm{d}t = X_t\left( (g(t)^2- a(t))Z_t\right)\mathrm{d}t\end{align*} let us conclude that $$ f_1(Z_t, t) = (-a(t)+g(t)^2)Z_t.$$ Thus $$\mathrm{d}Z_t = (-a(t)+g^2(t))Z_t \mathrm{d}t - g(t)Z_t\mathrm{d}B_t,$$ you can see the explicit form of $Z_t$ in Case 1 and $$\mathrm{d}(Z_tX_t) = (b(t)-h(t)g(t))Z_t\mathrm{d}t + h(t)Z_t\mathrm{d}B_t.$$

Using Ito's lemma we can show that $(Y_t):= (Z_t^{-1})$ is an Ito process such that it satisfies $$ \mathrm{d}Y_t = a(t)Y_t \mathrm{d}t + g(t)Y_t\mathrm{d}B_t,\ \ Y_0 = 1$$ it has an explicit form $$ Y_t = \exp\left(\int_0^t (a(s)-\frac{1}{2}g^2(s)) \mathrm{d}s +\int_0^t g(s)\mathrm{d}B_s\right).$$

Finally, we see that \begin{align*}\mathrm{d}(Z_tX_t) =& (b(t)-h(t)g(t))Z_t\mathrm{d}t + h(t)Z_t\mathrm{d}B_t \end{align*} yields \begin{align*}Z_tX_t =& Z_0X_0 + \int_0^t (b(s)-h(s)g(s))Z_s\mathrm{d}s + \int_0^t h(s)Z_s\mathrm{d}B_s \end{align*} Hence, we have the explicit solution \begin{align*}X_t =& Z_0X_0Y_t + Y_t\left(\int_0^t (b(s)-h(s)g(s))Z_s\mathrm{d}s + \int_0^t h(s)Z_s\mathrm{d}B_s\right). \end{align*}

[Math] Ito’s product rule: Three processes

I found the answer myself after some trying:

Again, Ito's product rule; $d(X_tY_t) = X_tdY_t + Y_tdX_t + dX_tdY_t$.

Substituting in for $P_t = X_tY_t$, we can write the SDE of the 3 processes as:

\begin{align} d(P_tZ_t) &= P_tdZ_t + Z_tdP_t + dP_tdZ_t\\ \text{(substituting back in ${P_t}$)}&= X_tY_tdZ_t + Z_td(X_tY_t) +d(X_tY_t)dZ_t\\ &= \underbrace{X_tY_tdZ_t + X_tZ_tdY_t + Y_tZ_tdX_t}_{C_t} + Z_tdX_tdY_t + d(X_tY_t)dZ_t\\ &= C_t + Z_tdX_tdY_t + X_tdZ_tdY_t + Y_tdZ_tdX_t + dX_tdY_tdZ_t\\ &=C_t + \sigma_{Y,t}\sigma_{Z,t}X_tdt + \sigma_{X,t}\sigma_{Z,t}Y_tdt + \sigma_{X,t}\sigma_{Y,t}Z_tdt \end{align}

Note: Neglecting all terms smaller than $dt$. However, I am still having trouble with the quotient $d(X_t/Y_t)$. Any help regarding this problem would be much appreciated.