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Interpreting Merton’s Self-Financing Equation with Stochastic Integrals

Article Quant Q&A · Author: user2520938

Summary

The answer explains Merton’s passage from discrete portfolio changes to continuous-time notation by rewriting the sums as integrals. It identifies the consumption term as an ordinary time integral, the left-endpoint sum as an Itô integral, and the sum of products of increments as quadratic covariation. This makes clear why individual price increments do not simply converge to zero: their accumulated products with other increments can converge to a nonzero process.

The answer also relates right-endpoint sums to a Stratonovich integral and states the conversion between Stratonovich and Itô forms, including the half-quadratic-covariation correction. Its explanation relies on stochastic-process integral conventions and suitable convergence assumptions; the informal discrete expressions alone do not provide a complete proof. It is a conceptual translation of the equations rather than a full derivation of the conditions under which each limit exists.

Key ideas

  • Merton’s discrete consumption sums correspond to an ordinary integral over time.
  • Left-endpoint sums of portfolio value times wealth increments correspond to an Itô integral.
  • Sums of paired increments converge to quadratic covariation under appropriate assumptions.
  • Right-endpoint sums can be expressed using a Stratonovich integral and an Itô correction.
  • Continuity of a price path does not imply that accumulated products of increments vanish.

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# Question on Merton's self financing derivation


# Question on Merton's self financing derivation












I'm reading Merton's Optimum Consumption and Portfolio Rules in a Continuous-time Model, and don't understand the step where he goes from discrete to continuous time. Specifically, my confusion is about the snippet below.

By the footnote, $P(t)$ is assumed to be right-continuous. If so, then $\lim_{h\to 0}P(t+h)=P(t)$, which is also used to go from (10) to (10'). However, would that not also imply that $\lim_{h\to 0}P(t+h)-P(t)=0$? If so, then I do not understand what's going on when going from (9) to (9'), where it is suggested that $\lim_{h\to 0}P(t+h)-P(t)=dP$ rather than $0$.

Another way to phrase my confusion would be: it seems that applying the limit process to the right hand side of the first equality sign in (9) produces a different result then what Merton suggests in (9').

Moreover, it seems that the limit $\lim_{h\to 0}C(t+h)h=0$, not $C(t)dt$. If mistaken, please enlighten me with a rigorous definition of the symbols involved that justifies the equality Merton mentions.

## Answer by Kurt G. (score 7)

https://quant.stackexchange.com/a/74897

Having to use a typewriter in 1970 Merton tried to find a notation that is as intuitive as possible at the risk of looking unrigorous at first glance. Since the advent of LaTeX it is easy to transcribe his formulas using Riemann-Stieltjes notation as follows (for simplicity I omit the subscripts $i$ in $N$ and $P$ and I won't take the sum over $i\,$):

In the limit, the LHS of Merton's equation (9) can be transcribed in integral form as

$$\tag{a} -\int_0^t C(s)\,ds\,. $$ The second line in Merton's equation (9) is the discrete (and differential) analogue of $$\tag{b} \lim_{m\to\infty\atop\|\Pi_m|\|\to 0}\sum_{j=1}^{m-1} \big[N(t^m_{j+1})-N(t^m_j)\big]\big[P(t^m_{j+1})-P(t^m_j)\big] $$ where $0=t^m_1<t^m_2\dots<t^m_m=t$ and $\|\Pi_m\|:=\max\limits_{j=1,\dots,m-1}|t^m_{j+1}-t^m_j|\,.$ Today we call (b) the quadratic covariation of $N$ and $P$ (see [1]) and denote it by $\langle N,P\rangle_t\,.$ A still popular notation (esp. among physicists) for the differential $d\langle N,P\rangle_t$ of (b) is $dN\,dP\,.$

The third line of (9) becomes $$\tag{c} \lim_{m\to\infty\atop\|\Pi_m|\|\to 0}\sum_{j=1}^{m-1} \big[N(t^m_{j+1})-N(t^m_j)\big]P(t^m_j) $$ which is the Ito integral $\int_0^t P(s)\,dN(s)\,.$

The first line in Merton's equation (9) is $$\tag{d} \lim_{m\to\infty\atop\|\Pi_m|\|\to 0}\sum_{j=1}^{m-1} \big[N(t^m_{j+1})-N(t^m_j)\big]P(t^m_{\color{red}{j+1}})\,. $$ With a bit of manipulation we can rewrite this as \begin{align} &2\lim_{m\to\infty\atop\|\Pi_m|\|\to 0}\sum_{j=1}^{m-1} \big[N(t^m_{j+1})-N(t^m_j)\big] \frac{P(t^m_{j+1})+P(t^m_j)}{2}\\ &~~-\lim_{m\to\infty\atop\|\Pi_m|\|\to 0}\sum_{j=1}^{m-1} \big[N(t^m_{j+1})-N(t^m_j)\big] P(t^m_j)\tag{e} \end{align} which we recognize as $$\tag{f} 2\int_0^t P(s)\color{red}{\circ}\,dN(s)-\int_0^t P(s)\,dN(s) $$ where the first one is a Stratonovich integral and the second again the Ito Integral.

Merton's chain of equations (9) can now be written as \begin{align}\tag{g} -\int_0^t C(s)\,ds&=2\int_0^tP(s)\circ dN(s)-\int_0^tP(s)\,dN(s)\\ &=\langle N,P\rangle_t+\int_0^t P(s)\,dN(s)\,. \end{align} Corrolary. Merton has (just in a few lines and using a 1970's typewriter) proven the equation $$\tag{h} \int_0^t N(s)\circ dP(s)=\int_0^t N(s)\, dP(s)+\frac{1}{2}\langle N,P\rangle_t\,. $$ (cf. Wikipedia).

I think he should have received the Nobel prize even if he had done nothing else than that.

The treatise of Merton is worth reading because his explanations of what is going on economically in these equations is marvellous.

[1] S. Shreve, Stochastic Calculus for Finance.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.