This is a very basic question relating to why we are allowed to multiply by random variables within an SDE. Every text/notes set that I've seen does the following, for $X_t$ continuous + adapted and $L_t$ a semimartingale:

$$dX_t = \frac{1}{X_t}dL_t \quad \text{as} \textbf{ shorthand} \text{ for meaning} \quad X_t = \int_0^t\frac{1}{X_s}dL_s \quad \quad (1)$$

Because it's only ever referred to as shorthand, I don't understand why no one ever proves the fact that we are allowed to then write: $$X_tdX_t = dL_t, \quad \text{i.e. once again shorthand for} \quad L_t = \int_0^tX_sdX_s \quad \quad (2)$$

What is the rigorous justification for $(2)$? Am I an idiot for missing something trivial about the definition of the integral or of Ito's formula that makes the equivalence of $(1)$ and $(2)$ obvious? Sorry if this is dumb. Please help if you can


This is my attempt, please let me know if you find some mistake as I am not 100% sure about some of calculations. ( I am probably missing a lot of rigorous steps, it's more like a sketch of a proof)

I start by writing

$$\int_0^t X_sdX_s=\int_0^t X_s d\left(\int_0^s X_u^{-1} dL_u\right)$$ Under certain conditions (assume $E(X_sX_t)$ is a continuos function of $t$ and $s$, otherwise we should use another approximating sequence of functions) the Ito integral can be seen as the followin limit (in $L^2$):

$$\int_0^t X_s d\left(\int_0^s X_u^{-1} dL_u\right)= \lim_{\|\Delta\|\to 0} \sum_{i}X_{t_i-1}\left[\left(\int_0^{t_i} X_u^{-1} dL_u\right)-\left(\int_0^{t_{i-1}} X_u^{-1} dL_u\right)\right]$$

$$=\lim_{\|\Delta\|\to 0} \sum_{i}X_{t_i-1}\left[\left(\int_{t_{{i-1}}}^{t_i} X_u^{-1} dL_u\right)\right]$$

$$=\lim_{\|\Delta\|\to 0} \sum_{i}X_{t_i-1} \left(\lim_{\|\Pi\|\to 0} \sum_{j} X^{-1}_{s^i_{j-1}}(L_{s^i_j}-L_{s^i_{j-1}})\right)$$

$$=\lim_{\|\Delta\|\to 0} \sum_{i} \left(\lim_{\|\Pi\|\to 0} \sum_{j} \color{red}{X_{t_i-1}X^{-1}_{s^i_{j-1}}}(L_{s^i_j}-L_{s^i_{j-1}})\right)$$

Since $X_{t_{i-1}}\leq X_{s^i_{j-1}}\leq X_{t_i}$ and the mesh is converging to zero, then they converge to the same value and the red term goes to $1$.

And then we end up with a telescopic sum which results in $L_t-L_0$.


In addition to the other answer, you can also see this by intrinsic properties of the Ito integral with no need for an approximation by Riemann type sums. I will denote the Ito integral of $f$ against $X$ by $f \bullet M$ if $M$ is a continuous local martingale (this notation is more convenient for writing out certain properties) and will write $f \cdot A$ for the Riemann-Stieltjes integral if $A$ is of finite-variation. Also let $X_t = M_t + A_t$ be the semimartingale decomposition of $X$ so that $M$ is a local martingale and $A$ is of finite variation.

The Riemann-Stieltjes integral has the associativity property that $(fg) \cdot A = f \cdot(g \cdot A)$. I claim that the Ito integral also has this property.

This will suffice to prove the desired result since this then shows that the integral against a semimartingale has the same associativity property. In particular, it follows that $X_t dX_t = X_t \frac{1}{X_t} dL_t = dL_t$.

(Associativity of the Ito Integral): $(KH) \bullet M = K \bullet (H \bullet M)$

Proof: This follows from associativity of the Riemann-Stieltjes integral since $K \bullet M$ is the unique continuous local martingale such that $$\langle K \bullet M, N \rangle_t = (K \cdot \langle M,N \rangle)_t$$ for all continuous local martingales $N$, where $\langle \cdot, \cdot \rangle_t$ is the quadratic variation. Indeed, we then have $$\langle K \bullet ( H \bullet M), N \rangle_t = \left( K \cdot \left( H \cdot \langle M,N\rangle \right) \right)_t = \left(KH \cdot \langle M,N \rangle \right)_t$$ so that $KH \bullet M = K\bullet(H \bullet M)$ by uniqueness.


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