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In Chapter 9 of Shreve's book Stochastic Calculus for Finance II, the main theorem is the 9.2.1. Defining the discounting process $D(t)=\mathrm{e}^{-\int_0^t du r(u)}$ and $r(u)$ the, possibly stochastic, interest rate process; the theorem states that any positive price process $N$, under the risk neutral measure with Brownian motion $\tilde{W}_t$, obeys the following equation: $$(i) d(D(t)N(t))=D(t)N(t)\nu(t)\cdot d\tilde{W}_t, $$ where $\nu(t)$ is defined using the Martingale Representation Theorem. Moreover, massaging the equation we can rewrite equation (i) as: $$(ii) dN(t)=r(t)N(t)dt+N(t)\nu(t)\cdot d\tilde{W}_t.$$ In the same chapter in section 9.4 he defines the zero-coupon bond as $$ (iii) B(t,T)=\tilde{E}[D(T)/D(t)|\mathcal{F}_t]=\tilde{E}[\mathrm{e}^{-\int_t^T du r(u)}|\mathcal{F}_t], $$ where $B(T,T)=1$ and the expectation is over the risk neutral measure. Clearly, since $B$ is defined as an expectation value it is a non-random function. Then, in the following, in order to define the T-forward measure, the author uses theorem 9.2.1 to write equation (i) for the zero-coupon bond as: $$(iv)d(D(t)B(t,T))=D(t)B(t,T)\sigma(t,T)\cdot d\tilde{W}_t$$ Now since the theorem applies also equation (ii) it is supposed to make sense and then: $$(v)dB(t,T)=r(t)B(t,T)dt+B(t,T)\sigma(t,T)\cdot d\tilde{W}_t.$$ I.e. the zero-coupon bond price is described by a stochastic Ito equation!

Now my problem is that since equation (iii) defines $B$ as a deterministic function,is it possible to write equation (v) for $B$? It makes sense that equation (iv) is an Ito equation since $D(t)$ is, in principle, a random function, but what is the meaning of equation (v) then? A possible justification could be constructed considering that, actually, in equation (v) the stochastic interest rate $r$ and the volatility process $\sigma$ given by the Martingale Representation theorem are connected to each other, and then it would be possible (in theory) to "simplify" equation (v) and obtain a deterministic differential equation describing $B$ as a function of $t$. Is this the correct way to look at equation (v)?

Or maybe I am wrong and $B(t,T)$ is a stochastic function. Then why an expectation value is not deterministic function?

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$B(t, T)$ (and in general conditional expectations of random variables, under the usual conditions) are not deterministic but $F(t)$ measurable. $r(t)$ here is also $F(t)$ measurable.

One intuitive way of explaining conditional expectations being $F(t)$ measurable but not deterministic: suppose a stock is worth \$100 on Jan 1 2024. On Jan 1 2024, we may expect it to be worth $105 on Jul 1 2024, then worth \$110 on Jan 1 2025. However, if at Jul 1 2024 it's reached a value of \$150, then we will likely adjust our expectation of the stock's price on Jan 1 2025 to be higher than \$110.

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  • $\begingroup$ Thank you for the answer. I am still confused though. A conditional expectation is just what is the average value of a stochastic function given the information at time t. I.e. if I were to average over all possible futures what should I expect. Now if the specific path diverges in the future at t+dt, like in your stock example, doesn't mean that the conditional expectation value is random variable. Perhaps with your example you meant that as time goes on the expectation value should be updated in view of the new information? $\endgroup$
    – apelle
    Jan 22 at 16:26
  • $\begingroup$ Your comment "Perhaps with your example you meant that as time goes on the expectation value should be updated in view of the new information?", I agree with. This is the nature of conditioning on a filtration. In particular, using terminology loosely here, it implies that given $s < t < T$, then $E[ X(T) | F(t) ]$ is random at time $s$ but "decided" and therefore nonrandom at time $t$. In the example of this question, this means $B(t, T)$ is a random variable when looking from time $s$ but not from time $t$. $\endgroup$
    – Rylan
    Jan 22 at 16:47
  • $\begingroup$ I am not sure I understand what you mean. $E[X(s)|F(t)]=X(s)$ for $s<=t$, because it is known. Then for $s>t$ $E[X(s,T)|F(t)]=B(t,T)$ that is a deterministic function. Think about a Brownian Motion $X$ that at time $t$ has a value of 4, then $E[X(s)|F(t)]=X(s)$ for $s<t$ and $E[X(s)|F(t)]=4$ for $s>t$. Which is the random part? Now of course if at time $u>t$ we would observe $X=-3$ we would have $E[X(s)|F(u)]=-3$ for $s>u$. $\endgroup$
    – apelle
    Jan 22 at 17:01
  • $\begingroup$ Thinking a bit more about it. Maybe equations (iv) and (v) describe how the expectation value $B(t,T)$ evolves has I measure/compute the integral implied by $E$ as time increases and the information/filtration is updated. Is this correct? $\endgroup$
    – apelle
    Jan 22 at 17:13
  • $\begingroup$ I think that makes sense (ofc we are using intuitive terminology not the technical stuff, so it's worth re-reading the definitions to ensure that your intuition aligns with the definitions and what you've seen so far) $\endgroup$
    – Rylan
    Jan 22 at 18:07

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