Ordinary calculus has the product rule d(xy) = x\,dy + y\,dx, and
its integral twin, integration by parts. In the stochastic world both pick up one extra
term — the
quadratic covariation
of the two factors. For Itô processes X, Y,
d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X, Y]_t,
and, integrating from 0 to T,
\int_0^T X_t\,dY_t = X_T Y_T - X_0 Y_0 - \int_0^T Y_t\,dX_t - [X, Y]_T.
The lone correction d[X, Y]_t is the entire difference from the
rule you learned in first-year calculus — and it is a direct application of
Itô's lemma
to the function f(x, y) = xy.
Deriving the product rule, line by line
Apply the two-variable Itô lemma to f(x, y) = xy. The two-variable
form, for processes X, Y, reads
df = f_x\,dX + f_y\,dY + \tfrac{1}{2}\Big(f_{xx}\,d[X] + 2f_{xy}\,d[X,Y] + f_{yy}\,d[Y]\Big),
the second-order part now carrying the covariation cross-term
2f_{xy}\,d[X,Y] alongside the two pure variations.
Step 1 — compute the partials of f(x, y) = xy:
f_x = y, \qquad f_y = x, \qquad f_{xy} = 1, \qquad f_{xx} = 0, \qquad f_{yy} = 0.
Step 2 — substitute. The two pure second derivatives vanish, so the variation
terms d[X] and d[Y] drop out entirely;
only the cross term remains:
d(X_t Y_t) = y\,dX_t + x\,dY_t + \tfrac{1}{2}\cdot 2\cdot 1\cdot d[X, Y]_t.
Step 3 — write x = X_t,
y = Y_t and simplify the \tfrac12\cdot 2 = 1:
d(X_t Y_t) = Y_t\,dX_t + X_t\,dY_t + dX_t\,dY_t.
Step 4 — recognise the cross-product as a covariation. From the box algebra,
dX_t\,dY_t = d[X, Y]_t, which delivers the product rule:
\boxed{\,d(X_t Y_t) = Y_t\,dX_t + X_t\,dY_t + d[X, Y]_t.\,}
Integrating both sides over [0, T] and moving one integral across
gives the integration-by-parts formula stated above. The extra
-[X, Y]_T is the Itô tax on the classical rule.
A worked example: d(t\,W_t) and \int_0^T t\,dW_t
Take X_t = t (pure drift, dX_t = dt) and
Y_t = W_t (dY_t = dW_t). The covariation
term d[t, W]_t = 0, because t has
finite variation — it is smooth, so it accumulates no covariation with
anything (formally its diffusion coefficient b^X = 0). Then
d(t\,W_t) = t\,dW_t + W_t\,dt + \underbrace{d[t, W]_t}_{=\,0} = W_t\,dt + t\,dW_t.
This one has no correction — the ordinary product rule already holds, precisely
because one factor is smooth. Now integrate from 0 to
T (using 0\cdot W_0 = 0) and rearrange to
evaluate an otherwise-awkward Itô integral:
T\,W_T = \int_0^T W_t\,dt + \int_0^T t\,dW_t \;\;\Longrightarrow\;\; \int_0^T t\,dW_t = T\,W_T - \int_0^T W_t\,dt.
Integration by parts has traded an integral against dW (a
stochastic integral) for one against ordinary dt — exactly the kind
of simplification that makes the formula a workhorse in pricing.
For Itô processes X, Y:
-
Product (differential) form:
d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X, Y]_t.
-
Integration-by-parts (integral) form:
\int_0^T X_t\,dY_t = X_T Y_T - X_0 Y_0 - \int_0^T Y_t\,dX_t - [X, Y]_T.
-
The extra d[X, Y] term vanishes whenever one
factor has finite variation (e.g. X_t = t), recovering the
ordinary product rule.
Why does the correction disappear for a smooth factor? In the cross-product
dX_t\,dY_t, a finite-variation process contributes a
dt in place of a dW, and a product with
a dt is always second-order
(dt\,dW = 0, (dt)^2 = 0). So the
covariation term is non-zero only when both factors carry genuine Brownian
roughness — the ordinary product rule fails exactly when both legs wiggle.
As a consistency check, set X_t = Y_t = W_t in the product rule.
Then d[W, W]_t = dt, and
d(W_t^2) = W_t\,dW_t + W_t\,dW_t + d[W, W]_t = 2W_t\,dW_t + dt,
which is exactly the d(W_t^2) we got straight from
Itô's lemma.
The product rule and the single-variable lemma agree, as they must — the
+\,dt is the covariation [W, W]_t = t
showing up once more.