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.

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  • 2
    $\begingroup$ If you are not comfortable with the way Merton wrote this I recommend to learn from a standard stochastic calculus text what quadratic (co)variation is. We have moved on and do not have to do it without more modern and more rigorous tools. $\endgroup$
    – Kurt G.
    Commented Mar 8, 2023 at 21:15
  • $\begingroup$ @KurtG. Can you give a reference to a rigorous book that deals with this issue? $\endgroup$ Commented Mar 9, 2023 at 7:17
  • $\begingroup$ BTW it is lovely to see typewritten math from 40 to 50 years ago. Where did you get this gem from? Regarding your question: Steven Shreeve, Stochastic Calculus for Finance should take you a very long way. $\endgroup$
    – Kurt G.
    Commented Mar 9, 2023 at 8:53
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    $\begingroup$ I think the questions @user2520938 is a legit one. And saying this was derived by a nobel prize so it must be right is not enough. Either the question is a "basic financial question" or it is not. If it is a basic financial question someone should be able to quickly comment the solution, if it is not it should remain reopened. The advice of "read Shreve" is not helpful to user2520938 who has a very specific question. $\endgroup$
    – phdstudent
    Commented Mar 9, 2023 at 19:25
  • 4
    $\begingroup$ I believe there is enough to discuss about these limits for this to have an interesting answer. $\endgroup$
    – Bob Jansen
    Commented Mar 9, 2023 at 21:45

1 Answer 1


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.

  • $\begingroup$ Thanks a lot! I'll read this once I have time and accept the answer. $\endgroup$ Commented Mar 11, 2023 at 17:30

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