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How to Choose an Itô Transformation to Solve an SDE

Original question

How do I solve stochastic differential equations using Itô's lemma, and can you walk me through an example?
Look for a change of variable that simplifies the drift, diffusion, or both. Apply Itô's lemma to the transformed variable, solve its new equation, and invert the transformation. Linear equations often respond to an integrating factor, while multiplicative noise of the form σXt dWt\sigma X_t\,dW_t suggests a logarithm or a stochastic exponential. Exact transformations work for particular structures. Numerical methods are useful when those structures are absent.

Choose a transformation by what it cancels

A good transformation makes a difficult SDE look like a familiar integral. In an ordinary linear equation, an integrating factor cancels the term proportional to the unknown. An Itô SDE with multiplicative noise has a stochastic analogue. The homogeneous solution carries the same growth and shocks as the full equation. Dividing by that solution isolates whatever is left, provided the inverse and product rules include the covariation correction.

A stochastic integrating factor, step by step

Start with the equation you would get by setting c=0c=0. Its solution ZtZ_t tells you how one unit of the initial balance evolves without deposits. Dividing by ZtZ_t should cancel that homogeneous motion, just as dividing by an ordinary integrating factor does in a linear ODE. The catch is that both XX and the inverse of ZZ have Brownian terms. Their product has a drift contribution from quadratic covariation, which is needed for the cancellation below.

Consider dXt=(c+μXt) dt+σXt dWtdX_t=(c+\mu X_t)\,dt+\sigma X_t\,dW_t with deterministic X0=xX_0=x and constant coefficients. The homogeneous part has solution

Zt=exp⁡ ⁣((μ−12σ2)t+σWt),Z0=1.Z_t=\exp\!\big((\mu-\tfrac12\sigma^2)t+\sigma W_t\big),\qquad Z_0=1.

Verify this by applying Itô's lemma to the exponential. To remove the homogeneous part, write Yt=Zt−1XtY_t=Z_t^{-1}X_t. The inverse satisfies

d(Zt−1)=(σ2−μ)Zt−1 dt−σZt−1 dWt.d(Z_t^{-1})=(\sigma^2-\mu)Z_t^{-1}\,dt-\sigma Z_t^{-1}\,dW_t.
The product rule has three pieces. The two ordinary terms give cZt−1 dt+σ2Zt−1Xt dtcZ_t^{-1}\,dt+\sigma^2 Z_t^{-1}X_t\,dt, with the stochastic terms cancelling. The covariation contributes −σ2Zt−1Xt dt-\sigma^2Z_t^{-1}X_t\,dt, leaving dYt=cZt−1 dtdY_t=cZ_t^{-1}\,dt. Integrate and multiply by ZtZ_t:
Xt=Zt(x+c∫0tZs−1 ds).X_t=Z_t\left(x+c\int_0^t Z_s^{-1}\,ds\right).
This representation is explicit in the Brownian path. The time integral is random and can be evaluated numerically from a simulated path. To check the solution, apply the product rule to ZtZ_t and the expression in parentheses. The parenthesized factor has finite variation, so their covariation vanishes.

What the integrating factor does to each deposit

Suppose a stylized account starts at 100100, receives deposits continuously at c=12c=12 units per year, and its invested balance has drift μ=0.05\mu=0.05 and volatility σ=0.20\sigma=0.20. Rewrite the solution as Xt=100Zt+12∫0t(Zt/Zs) dsX_t=100Z_t+12\int_0^t (Z_t/Z_s)\,ds. The ratio Zt/ZsZ_t/Z_s is growth earned by a unit deposited at time ss. Its expectation is e0.05(t−s)e^{0.05(t-s)} by the normal moment-generating function. At one year the expected balance is 100e0.05+12(e0.05−1)/0.05100e^{0.05}+12(e^{0.05}-1)/0.05, or approximately 117.43117.43. With deposits and a positive initial balance, the exact solution stays positive because the exponential and the parenthesized term are both positive.

The same expectation follows by taking expectations in the original SDE and solving m′(t)=c+μm(t)m'(t)=c+\mu m(t). This provides a check on the integrating factor calculation.

References and next steps