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For an arbitrary semimartingale $Y_t$ show that $\mathcal{E}(\mathcal{L}(Y_t)) = Y_t/Y_0$.

I saw this post and tried to do the same for a non necessarily continuous process but can not really see the equality of the terms involving the jumps of the process. Hence I tried to use the other characterisation of the stochastic logarithmic and the exponential functions in terms of their differentials. Namely, that $W_t = \mathcal{E}(Y_t)$ is the process that satisfies $dW_t = W_t dY_t$ and if $W_t = \mathcal{L}(Y_t)$ then it satisfies $dW_t = dY_t/Y_t$.

Denoting $X_t = \mathcal{E}(\mathcal{L}(Y_t))$, using this differential characterisation it is easy to check that:

$$\frac{dX_t}{X_t} = \frac{dY_t}{Y_t}$$

or that $\mathcal{L}(Y_t) = \mathcal{L}(X_t)$ but possibly differing at their initial point. Can we conclude from this our claim? Can anybody add some comments on it?

Barreto
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1 Answers1

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The linked post defines neither the stochastic exponential, nor the stochastic logarithm of a semi-martingale, henceforth abbreviated by SM:

  • The stochastic (aka Doleans-Dade) exponential of a SM $X_t$ is the unique strong solution to the SDE $$ dY_t=Y_{t-}\,dX_t\,, \quad Y_0=1\, $$ and is denoted by $Y_t={\cal E}(X_t)\,.$

  • The stochastic logarithm of the SM $Y_t$ is the SM $X_t$ given by $$ dX_t=\frac{dY_t}{Y_{t-}}\,,\quad X_0=0\, $$ and is denoted by $X_t={\cal L}(Y_t)\,.$

The claim $$\tag{1} \boxed{\phantom{\Big|}\quad {\cal E}({\cal L}(Y_t))=Y_t\quad} $$ is now trivial by plugging the definitions into each other.

In your notation \begin{align} X_t&={\cal E}(W_t)\,,\quad &dX_t&=X_{t-}\,dW_t\,,\;&X_0=1\,,\\[2mm] W_t&={\cal L}(Y_t)\,,\quad &dW_t&=\frac{dY_t}{Y_{t-}}\,,\;&W_0=0\,. \end{align} In particular $$ dY_t=Y_{t-}\,dW_t\,. $$ It follows that $X_t=Y_t$ holds as soon as we require $Y_0=1$ because that SDE has a unique strong solution. This implies again (1).

Kurt G.
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  • Thanks for the answer. But it is a little confusing since you are already seting $Y_t = \mathcal{E}(\mathcal{L}(Y_t))$, but did is what we have to show. – Barreto Apr 16 '23 at 11:45
  • @Sarah I copied the definitions I was missing in your and the linked post from Wikipedia. My answer is self-contained and there is nothing to show. If this is not what you wanted I am awaiting an edit to your question giving other definitions of ${\cal E}$ and ${\cal L},.$ – Kurt G. Apr 16 '23 at 11:48
  • @Sarah Having said that: when I read OP again word for word your defintions are pretty close. Just remember where those processes start and use perhaps another notation: $Y_t={\cal L}(W_t)$ should do. – Kurt G. Apr 16 '23 at 11:53
  • I think it would be more clear if we let $W_t = \mathcal{L}(Y_t)$ and $X_t = \mathcal{E}(W_t)$. But then it is not so obvious to me that $X_t = Y_t$ (being not carefull with the initial positions). – Barreto Apr 16 '23 at 12:22
  • @Sarah Addressed that in an edit. I would strongly recommend to be careful with the initial positions. – Kurt G. Apr 16 '23 at 12:47
  • Thank you, crystal clear now! – Barreto Apr 16 '23 at 13:24