Tag Archives: shreve vol. I

Shreve I-6.2: Risk-neutral measure

Tokyo update
Suppose we adopt the axiom that $\mathbb E(S_1)=(1+\epsilon)S_0$ where $\epsilon\ne r$. This is similar to a risk-neutral outlook except that, well, it is no longer risk-neutral.

Since $V_1$ and $S_1$ depend on head vs. tail, we have $V_1=\alpha S_1+\beta$ for some constants $\alpha$ and $\beta$ that can be calculated. The relationship between $V_1$ and $V_0$ has the property that
$$
E(V_1/V_0)=E(V_1)/V_0=\frac{\alpha E(S_1)+\beta}{\alpha S_0+\frac{\beta}{1+r}}
$$
since a bond is still risk-free hence risk-neutral when the interest rate is known.
$$
=\frac{\alpha (1+\epsilon)S_0 +\beta}{\alpha(1+r)S_0+\beta}\cdot (1+r)
$$
When $\epsilon=r$ this simplifies to $1+r$; in particular then $E(V_1/V_0)$ is the same for all securities $V$ when $\epsilon=r$. For any value of $\epsilon\ne r$, $E(V_1/V_0)$ depends on $\alpha$ and $\beta$.

Can we still find $V_0$ if $\epsilon\ne r$? Yes, the value of $V_0=\alpha S_0+\frac{\beta}{1+r}$ does not depend on $\epsilon$. But using $\epsilon\ne r$, we cannot relate $V_0$ to $E(V_1)$ in a simple way.

Updated September 24.

Each measure has the property that if for a random variable $X$, $\tilde E(X)=0$, and $\tilde P(X>0)>0$, then $\tilde P(X < 0)>0$.

In the study of interest rates, the so-called risk-neutral measure can to some extent be simply a fixed measure giving positive probability to each outcome.
Such a measure of course has the property that if for a random variable $X$, $\tilde E(X)=0$, and $X(\omega)>0$ for some $\omega\in\Omega$,
then $X(\omega) < 0$ for some $\omega\in\Omega$.

Now let us consider another measure $P$ which is the “real” probability, and which also gives positive probability to each outcome.

Then we have that if for a random variable $X$, $\tilde E(X)=0$, and $P(X>0)>0$, then $P(X < 0)>0$.
So if $X$ is the outcome of some trading strategy, then if we can verify that $\tilde E(X)=0$ then we will have ruled out arbitrage.
Now we can use $\tilde P$ and $\tilde E$ to define the “fair” prices of assets, in such a way that $\tilde E(X)=0$ follows.
Then these prices will indeed be fair, or at least arbitrage-free.

It does not follow that $\tilde P$ is in any sense “risk-neutral”, however. In Chapters 1–5 we had a stock valued at $S_0$ and moving to a future value $S_1(\omega_1)$. Then we defined the risk-neutral measure by the axiom $(1+r)S_0=\tilde E(S_1)$ (the equation that expresses risk-neutrality). That made sense because any other price $\hat S_0$ for the stock at time 0 would lead to an arbitrage: the stock would be trading at $S_0$ but would really be worth $\hat S_0$! (It is as if you could buy a 5-dollar bill for 1 dollar.) But in Chapter 6, there is no asset whose price at time 0 is given, there are only the random interest rates, and so we just fix one measure $\tilde P$ to price with.

That being said, there could well be an ambient stock to determine a risk-neutral measure:
Namely, the interest rate $R_n$ applied during $[n,n+1]$ is already known at time $n$, so it is possible to determine a risk-neutral measure from the stock price paths. Conversely, from the strange risk-neutral measure given in the first example of Ch. 6, and from the interest rates added to that example, one can determine a set of stock price paths (Figure 6.3.1). This procedure is not unique, though. For instance,
$$
\frac{1+r-d}{u-d}=\tilde p=P(HHH | HH) = 2/3
$$
and $\tilde q = P(HHT | HH) = 1/3$. And the interest rate is $r=R_2(HH)=1$. This gives
$$
\frac{2-d}{u-d}=\tilde p=P(HHH) = 2/3
$$
$$
3(2-d)=2(u-d)
$$
$$
6=2u+d
$$
For instance $u=8/3$ and $d=2/3$.

And $P(HTH | HT) = 1/2$ with $R_2(HT)=0$, giving
$$
\frac{1+0-d}{u-d}=\tilde p=1/2
$$
$$
2(1-d) = u-d
$$
$$
u+d=2
$$
For instance, $u=3/2$ and $d=1/2$.

Conclusion: it does makes sense to call $\tilde P$ a risk-neutral measure in Ch. 6 on interest rates, even though its risk-neutrality is not necessary for its use in that chapter.

Shreve I-5: Distribution of first hitting time

Let $\tau$ be the first hitting time of 1 for a symmetric simple random walk where $p=q=1/2$. Then $$ P(\tau\le 2j-1)=2 P(S_{2j}\ge 1)= 2 P(S_{2j}\ge 2)= P(S_{2j}\ge 2 $$ or $$ S_{2j}\le -2) = 1 - P(S_{2j}=0) = 1 - {2j\choose j}p^j q^j $$ because after hitting 1 the walk may have gone either up or down. and so $$ P(\tau=2j-1) = {2j-2\choose j-1}p^{j-1}q^{j-1} - {2j\choose j}p^jq^j $$ $$ = 2^{-2j} \left( 4 \frac{(2j-2)!}{(j-1)!(j-1)!} - \frac{(2j)!}{j!j!}\right) $$ $$ = 2^{-2j} \frac{(2j-2)!}{j!(j-1)!} \left( 4j - (2j)(2j-1)/j\right) $$ $$ = 2^{1-2j} \frac{(2j-2)!}{j!(j-1)!} \left( 2j - (2j-1)\right) = 2^{1-2j} \frac{(2j-2)!}{j!(j-1)!} $$ If $p\ne q$ it is a little harder. We want to find the number $C_n$ of paths of length $2n$ starting and ending at zero that do not hit 1. This is ${2n\choose n}$ minus the number of paths starting and ending at 0 that do hit 1. This subtracted number, in turn, is the number of paths starting at 0, hitting 1, and ending at 2 (by flipping the part of the path that comes strictly after hitting 1 for the first time). This is of course just ${2n\choose n+1}$ because in order to end at 2 after $2n$ steps, there must be $n+1$ upticks and $n-1$ downticks. So $$ C_n = {2n\choose n} - {2n\choose n+1} = {2n\choose n}\left(1-\frac{n}{n+1}\right) = {2n\choose n}\frac{1}{n+1} $$ Hence $$ C_{j-1} = \frac{(2j-2)!}{(j-1)!(j-1)!}\frac{1}{j} = \frac{(2j-2)!}{(j-1)!j!} $$ and $$ P(\tau = 2j-1) = p^{j}q^{j-1} C_{j-1} $$ To find the moment generating function in the case $p=q=1/2$, $$ E(\alpha^\tau) = \sum \alpha^{2j-1} P(\tau=2j-1) = \sum \alpha^{2j-1} p^j q^{j-1} C_{j-1} $$ $$ = \sum_{j=1}^\infty (\alpha/2)^{2j-1} C_{j-1} $$ $$ = \sum_{t=0}^\infty (\alpha/2)^{2(t+1)-1} C_{t} $$ $$ = \frac{\alpha}{2} \sum_{t=0}^\infty (\alpha/2)^{2t} C_{t} $$ Note that the Catalan numbers satisfy $C_0=1$ and $$ C_{n+1}=\sum_{i=0}^n C_i C_{n-i} $$ roughly speaking because in a path starting at 0 and ending at 0, never hitting 1, there is a last time $i$ that the path hits 0 (before the very last time) and the very last step must be an uptick; and the ticks in between can be done in $C_i$ and $C_{n-i}$ ways respectively. Now define the generating function $c(x)=\sum_{n=0}^\infty C_n x^n$. The recurrence relation for $C_{n+1}$ looks "quadratic" so let us see if $c(x)$ satisfies a quadratic equation: We have $$ c(x)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty C_n C_m x^{n+m} $$ $$ = \sum_{s=0}^\infty x^s \sum_{t=0}^s C_t C_{s-t} $$ and therefore also $$ c(x)=C_0x^0+\sum_{n=1}^\infty \sum_{i=0}^{n-1} C_i C_{n-1-i} x^n $$ $$ = 1 + \sum_{n=1}^\infty \sum_{i=0}^{n-1} C_i C_{n-1-i} x^n $$ $$ = 1 + xc(x)^2 $$ We can now solve $c=1+xc^2$; namely $xc^2-c+1=0$ gives $$ c = \frac{1\pm\sqrt{1-4x}}{2x} $$ If $x=(\alpha/2)^2=\alpha^2/4$ then $$ c = \frac{1\pm\sqrt{1-\alpha^2}}{\alpha^2/2} $$ and $$ \frac{\alpha}{2} c = \frac{\alpha}{2} \frac{1\pm\sqrt{1-\alpha^2}}{\alpha^2/2} = \frac{1\pm\sqrt{1-\alpha^2}}{\alpha} $$ Because $\alpha=1$ gives $E(\alpha^\tau)=E(1^\tau)=1$, we choose $\pm = -$, giving finally $$ E(\alpha^\tau) = \frac{1-\sqrt{1-\alpha^2}}{\alpha} $$ This can be applied to get some information about the price a perpetual put (Shreve I-5.4). Namely consider a perpetual put with strike $K=4$ in the usual model ($S_0=4$, $u=2$, $d=1/2$). One strategy is to exercise the put option as soon as the stock reaches a particular value like $2^k$ where $\infty < k\le 1$. We can calculate the value of this strategy, i.e., the price of a modified option where one has to use this strategy to get a stopping time $\tau$. Namely, $$ V_0 = \tilde E\left( ({1+r})^{-\tau} (4-2^k)\right) = (4-2^k) \tilde E\left(({1+r})^{-\tau}\right) $$ The hard part here is to evaluate $\tilde E(\alpha^\tau)$ for a constant $\alpha$. This is nothing other than the moment generating function!

What about hitting 2?

To hit 2 means to first hit 1 and then hit again starting from the new vantage point. So $$ P(\tau_2 = 2k) = \sum_{j=1}^k P(\tau_1 = 2j-1) P(\tau_1 = 2k - (2j-1) = 2(k-j+1) - 1) $$ $$ = \sum_{j=1}^k p^{k+1}q^{k-1} C_{j-1}C_{k-(j-1)-1} $$ $$ = \sum_{j=1}^k p^{k+1}q^{k-1} C_{j-1}C_{(k-1)-(j-1)} $$ $$ = p^{k+1}q^{k-1} \sum_{t=0}^{k-1} C_{t}C_{k-1-t} $$ $$ = p^{k+1}q^{k-1} (C_{k}) $$

Shreve I-5: Recurrence and biased random walks

Here we give a proof that random walk is recurrent for $p=1/2$ and not for other values of $p$.

From Shreve’s older book “Stochastic calculus and financial applications” p. 6 we have:
Let $p$ be the probability of a step to the right, $X_i=+1$, and $q=1-p$.
Let $f(k)$ be the probability of reaching $A$ before $-B$ given that $S_0=k$. Then
$$
f(k)=pf(k+1)+qf(k-1)
$$
In terms of $\Delta f(k):=f(k+1)-f(k)$ this says
$$
\Delta f(k)=(q/p)\Delta f(k-1)
$$
so
$$
\Delta f(k+j)=(q/p)^{j}\Delta f(k)
$$
and so
$$
f(k)=f(-B)+ \sum_{i=-B}^{k-1} \Delta f(i)
= 0 + \sum_{i=-B}^{k-1} \Delta f(i) =
\sum_{j=0}^{k+B-1}\Delta f(j-B)
$$
$$
= \Delta f(-B) \sum_{j=0}^{k+B-1}(q/p)^j
$$
$$
= f(-B+1) \sum_{j=0}^{k+B-1}(q/p)^j
$$
$$
= f(-B+1) \frac{(q/p)^{k+B}-1}{(q/p)-1}
$$
Since $f(A)=1$ we can evaluate $f(-B+1)$ as $\frac{(q/p)-1}{(q/p)^{A+B}-1}$.
Then
$$
f(0)= \frac{(q/p)-1}{(q/p)^{A+B}-1} \frac{(q/p)^{B}-1}{(q/p)-1}$$
Thus

$$
P(S_n \text{ hits }A\text{ before }-B \mid S_0=0) = \frac{(q/p)^B-1}{(q/p)^{A+B}-1}
$$
Setting $A=B$ we have
$$
P(S_n \text{ hits }A\text{ before }-A \mid S_0=0) = \frac{(q/p)^A-1}{(q/p)^{2A}-1} = \frac{1}{(q/p)^{A}+1}
$$
Let us consider the probability of never hitting $-A$. This is the same as hitting $A$ before $-A$, then $3A$ before $-A$ (so in effect $2A$ before $-2A$), then $7A$ before $-A$ (so in effect $4A$ before $-4A$), and so on, so we get
$$
\prod_{k=1,2,4,\ldots}\frac{1}{(q/p)^{kA}+1}
=
\prod_{n=0}^\infty\frac{1}{(q/p)^{A2^n}+1}
$$
Let us set $A=2^a$ where $a\ge 0$, $a$ an integer (where $a=0$ is perhaps most interesting), so we are looking at
$$
\prod_{n=0}^\infty\frac{1}{(q/p)^{2^{n+a}}+1}
$$
If $p\le 1/2$ then $q/p\ge 1$ and so this is
$$
\le \prod_{n=0}^\infty\frac{1}{1^{2^{n+a}}+1}=0
$$
since the factors are bounded below 1. But if $p>1/2$ then $\alpha:=q/p<1$ and we have the equivalent conditions
$$
\prod_{n=0}^\infty\frac{1}{\alpha^{2^{n+a}}+1} > 0
$$
$$
\log\prod_{n=0}^\infty\frac{1}{\alpha^{2^n}+1} > -\infty
$$
$$
\sum\log\frac{1}{\alpha^{2^n}+1} > -\infty
$$
$$
\sum\log({\alpha^{2^n}+1}) < \infty
$$
But $\ln(x)\le x-1$ for all $x>0$, so
$$
\sum\log({\alpha^{2^n}+1}) < \sum \alpha^{2^n} < \sum\alpha^n < \infty
$$
Moreover as $a\rightarrow\infty$ the nonzero product actually approaches 1. Thus the probability that each negative number will eventually be reached is zero when $p>1/2$. Similarly, if $p<1/2$ then the probability that each positive number will eventually be reached is zero.

Shreve I-4 American call options

A non-path-dependent American derivative pays an amount $g(s)$ depending only on the current stock price $s$ whenever it is exercised.
The value at a current time is the maximum of the value of immediate exercise and the value of delayed exercise; the latter is defined recursively as before.
So if you have a choice between \$5 and a derivative valued at \$4, you should always choose the \$5, because in theory you could use \$4 to buy the derivative immediately and put the other \$1 in the bank. That is, we are assuming trading that we do is instantaneous whereas stock movements and interest accrual are not. If the stock could rise sharply while we are choosing the \$5, the logic would be different.

A path-dependent American derivative is really no different: we again price the derivative at any time as the maximum of the value of instantaneous exercise and the value of delayed exercise.

Note that the notion of optimal exercise of an option is straightforward, to determine it is merely a matter of choosing the largest of two numbers.

A stopping time in the $N$-period binomial model is a random variable taking integer values from $0$ to $N$ or $\infty$, and whose value $\tau(\omega)$ only depends on $\omega_1\cdots\omega_{\tau(\omega)}$. In other words, any choice of the remaining bits of $\omega$ gives the same value to $\tau$.

The notion of a stopping time can be used to express the price of an American derivative without talking about maxima of two numbers and recursion.
Namely, the price $V_n$ is the maximum over all stopping times (i.e., over all option exercise algorithms) taking values $\ge n$, of
$$
\tilde{\mathbb E}\left( 1_{\tau\le N} G_\tau/(1+r)^{\tau-n}\right)
$$
Here $G_n$ is the value of the derivative if exercised after $n$ steps.

In other words, price the derivative by the discounted risk-neutral expected payoff according to the exercise algorithm that maximizes it.

To justify the price formula, note that (1) once $\tau$ is fixed the option is equivalent to a European one, if we put payoff in the money market after exercise, and (2) if we assume that prices for options exist at every place in the tree then the optimal $\tau$ strategy is just to choose optimally (to exercise or not) at each stage.

We can note that a call option tends to gain value as time goes by, since we get to buy for a smaller and smaller amount in discounted terms. A put option tends to lose value, because being allowed to sell for 100 dollars is not so desirable in the future when 100 dollars is a smaller amount in discounted terms. Therefore a call option should be exercised at time $N$ (and so American call = European call, in effect). The textbook proves this by going via convex functions (a call option has a convex payoff function, and Jensen’s inequality for convex functions is used) but we can consider the call option case directly: $g(s)=(s-K)^+$. We want to show that
$$
\max_{\tau\in\mathcal S_n}\tilde{\mathbb E}\left( 1_{\tau\le N} g(S_\tau)/(1+r)^{\tau-n}\right) = \tilde{\mathbb E}\left( S_N^+ / (1+r)^N \right)
$$

Let us assume $S_n>K$ and compare exercise at time $n$ with waiting one more step.
Case 1: $uS_n$ and $dS_n$ are both $>K$. Then
$$
(1+r)^{-1} \tilde{\mathbb E}(S_{n+1}-K)^+ = (1+r)^{-1} ((1+r)S_n – K) = S_n – \frac{K}{1+r} \ge S_n – K.
$$
Case 2: Otherwise, then $dS_n<K$. We may assume $S_n> K$. At the next step we $\tilde{\mathbb E}$-expect the stock price to be at $(1+r)S_n$ and so
it is enough to check that
$$
(1+r)^{-1}\tilde{\mathbb E}(S_n-K)^+ = (1+r)^{-1} (uS_n – K) \tilde{\mathbb P}(H) = (1+r)^{-1}(uS_n-K)\frac{1+r-d}{u-d} \ge^{!} S_n – K
$$
We proceed as follows:
$$
S_n-K \le \frac{uS_n-K}{1+r}\cdot\frac{1+r-d}{u-d}
$$
$$
(1+r)(u-d)(S_n-K) \le (uS_n-K)(1+r-d)
$$
$$
(1+r)(u-d)(S_n-K) \le (uS_n-K)(1+r-d)
$$
$$
[(u-d-1)r+(u-1)]K \ge (u-1-r)dS_n
$$
For this it suffices, since $dS_n\le K$, that
$$
(u-d-1)r+(u-1) \ge u-1-r
$$
This is true since $u\ge d$.

Shreve I-1: Binomial asset pricing model

Wherein we give a slightly more intuitive version of the central replication derivation. Suppose we have a derivative security (which here really just means a random asset) worth $V_1(\omega_1)$ at time 1 and seek to determine its fair price at time 0, $V_0$. We will have $V_0=X_0$ where $X_0$ is an as-yet unknown amount of money that will be needed to replicate the security.

The security presumably depends, whether positively or negatively, on a stock valued at $S_t$ at time $t$. So to replicate the security we buy some as-yet unknown amount $\Delta_0$ of shares of the stock.

((The whole thing is easier in the case where interest rates are $r=0$. Say the security returns either 10 or 17 depending on whether the stock is at 5 or 2, respectively. Then we just seek to express the security as a linear function of the stock, i.e. we seek numbers $\Delta_0$ and $c$ such that
$$
\Delta_0(5\text{ or }2)+c = \text{10 or 17},
$$
and then value the security at $\Delta_0 S_0+c$.

If we find “risk-neutral” probabilities under which the stock has expected value equal to its current value (these must exist, i.e., we must have $dS_0<S_0<uS_0$, or else all money should be taken out of the money market and invested in the stock), then the price of the security is just the expected value of the security under these probabilities. [Proof: First, this is true if the security is just equal to the stock, by definition of risk-neutral probabilities. Then as we saw above a general security is a linear function of the stock, and expectations preserve linear combinations.] The main advantage of this is not that we believe in a risk-neutral world — we might as well have used a world where the risk premium $\mathbb E S_1/S_0$ is 10% or some other fixed easy-to-compute number (well, assuming we could be sure that $\tilde p$ and $\tilde q$ would exist in that case too!) (and, well, dividing 1 by is significantly easier than dividing by any other number) — but that once we have the risk-neutral probabilities we can calculate the prices of many securities as long as they are all based on the same underlying set of stocks.
))

This leaves us with the cash position, i.e. money on hand, of $X_0-\Delta_0 S_0$, which of course we invest in the money market, i.e., we let somebody (such as a bank) borrow the money in return for paying us interest.

At time 1 the value of our portfolio (of stock and money market accounts) is

$$
X_1(\omega_1) = \Delta_0 S_1(\omega_1) + (1+r)(X_0-\Delta_0 S_0)
$$
Here $\omega_1\in\{H,T\}$, so we actually have two equations in the two unknowns $\Delta_0$, $X_0$. We impose $V_1(\omega_1)=X_1(\omega_1)$ for both $\omega_1$, and we assume $V_1(\omega_1)$ is known for each $\omega_1$.

Rather than relying on magic intuition, we solve this system of equations, using the fact that the inverse of the matrix
$$
\left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right]
$$
is
$$
\frac{1}{ad-bc}
\left[\begin{array}{rr}
d & -b \\
-c & a \\
\end{array}\right]
$$
Only then is it time to bring in the risk-neutral probabilities $\tilde p$ and $\tilde q$. Namely, we are curious whether there exist probabilities so that the expected value of the stock is just the return from the money market. Under real-world probabilities this should not happen, since it would make it needlessly risky to invest in stocks.

Then it turns out that, lo and behold,

the risk-neutral-expected return of our derivative security is equal to the money market return of… our sought-for time 0 price of the security.

So from these risk-neutral probabilities we can calculate the value $V_0$ of the security, using $(1+r)V_0 = \tilde{\mathbb{E}} V_1$.

Shreve I.3: State prices and other option prices

Suppose the stock having current share price $S_0$ may either go up or down, to $S_1(H)=uS_0$ or $S_1(T)=dS_0$. Suppose we have a “stock option” $V_1$ that pays 1 dollar if $H$ occurs and zero otherwise; $V_1(H)=1$ and $V_1(T)=0$. By the binomial asset pricing model (Shreve volume I chapter 1) the price of this option should be
$$
V_0 = \frac{1}{1+r} (\tilde p V_1(H) + \tilde q V_1(T)) = \frac{\tilde p}{1+r}
$$
[This is the state price of the state H.] Here
$$
\tilde p = \frac{1+r-d}{u-d}
$$

Suppose we fix $d=1/2$ and let $u\rightarrow\infty$. So we are looking at a stock that will either drop by 50%, or increase enormously. Then $V_0\rightarrow 0$, i.e., the option becomes worthless.

This may seem surprising since the option depends on heads/tails but not explicitly on the stock price associated with H/T. But the way to look at it is this: if there exists a stock that will either drop by 50% or increase enormously, then it better be very unlikely that the stock should increase.

Say $u=1000$ and $r=1/2$. Then to replicate the option we would buy $\Delta_0 = \frac{V_1(H)-V_1(T)}{S_1(H)-S_1(T)} = \frac{1}{999.5 S_0}$ shares. Because of the $S_0$ in the denominator this is just $\epsilon:=\frac{1}{999.5}$ dollars worth of shares.

We would need an initial capital of
$$
X_0 = \frac{1+r-0.5}{(1000-0.5)(1+r)} = \frac{1}{(3/2)(999.5)} = 2\epsilon/3
$$
which is two-thirds of the amount we invested in the stock. With such initial capital we have set up a portfolio guaranteeing to get exactly the value of the option $V_1(\omega)$ back at time 1.

Case 1: Tails occurs. Then the value of the option is zero, and the value of our portfolio is calculated as follows: we had to borrow $\epsilon/3$ to buy stock, which means we now owe somebody $(1+r)\epsilon/3 = \epsilon/2$. But the stock we bought now has value $\epsilon/2$; so we have 0 overall.

Case 2: Heads occurs. The value of the option is 1; we owe somebody $\epsilon/2$ but we the stock we bought has value $1000\epsilon$; and indeed
$$
1000\epsilon – \epsilon/2 = 1
$$

Generally the portion of our capital that we buy stock for is
$$
\frac{\Delta_0 S_0}{X_0} = \frac{(\Delta V/\Delta S)S_0}{[\tilde p V_1(H)+\tilde q V_1(T)]/(1+r)}
= \frac{\Delta V/(u-d)}{[\tilde p V_1(H)+\tilde q V_1(T)]/(1+r)}
$$
$$
= \frac{\Delta V/(u-d)}{[\tilde p \Delta V + V_1(T)]/(1+r)}
$$

$$
= \frac{\Delta V/(u-d)}{[((1+r-d)/(u-d)) \Delta V + V_1(T)]/(1+r)}
$$

$$
= \frac{(1+r)\Delta V}{[((1+r-d)/(u-d)) \Delta V + V_1(T)](u-d)}
$$

$$
= \frac{(1+r)\Delta V}{(1+r-d)\Delta V + V_1(T)(u-d)}
$$

$$
= \frac{(1+r)\Delta V}{(1+r-d)V_1(H) – (1+r-u) V_1(T)}
$$

$$
= \frac{(1+r)\Delta V}{((1+r)-d)V_1(H) + (u-(1+r)) V_1(T)}
$$
where $\Delta V = V_1(H)-V_1(T)$. It is worth looking at when this fraction will be more than 1, i.e., replicating the option involves borrowing from the money market. This is when
$$
\frac{V_1(H)}{V_1(T)} > \frac{S_1(H)}{S_1(T)}
$$
i.e., the option payoff is, in some sense, more dependent on the heads/tails outcome than the stock price is.
For instance, if the option were constant ($V_1(H)=V_1(T)=c$) then we need not borrow to replicate it; we would buy 0 shares of stock and invest the whole initial capital $c/(1+r)$ in the money market.

Shreve II.6.5 Why is the yield curve increasing and concave down?

Link to stockcharts.com (Click on Animate!)

Suppose the probability of default during a year $t$, given that there was no default before year $t$, is $p(t)$. We are considering a loan with term $T$ and constant annual interest $r$, with a principal of 1 and no dividend payments. We stipulate that the expected money we get back at time $T$ should equal the principal. So
$$
1 = \mathbb E((1+r)^T \cdot 1_{\text{no default}}) = (1+r)^T \prod_{t=1}^T (1-p(t))
$$
This gives
$$
\ln(1+r) = \frac{1}{T} \sum_{t=1}^T (-\ln(1-p(t))
$$
We use the approximation $\ln (1+x)\approx x$ and get
$$
r \approx \frac{1}{T} \sum_{t=1}^T p(t).
$$
To use calculus let us instead write

$$
r(T) = \frac{1}{T} \int_{t=1}^T p(t) dt.
$$
Now $dr/dT>0$ if and only if $p(T)> \frac{1}{T}\int_{0}^T p(t)dt$, in other words $p(\cdot)$ is increasing. To use the approximation we must assume that $p(\cdot)$ nevertheless stays close to $0$ in any single year.

So the yield curve being increasing corresponds to an increasing risk of default per year. Now notice that $p(t)$ is bounded above, certainly by 1 and hopefully by something much less than 1.

But whenever $p(t)$ is an increasing and bounded function, then its running average $r(t)$ will be increasing and converging to $\lim_{t\to\infty}p(t)$. Hence $r(t)$ will increase by less and less as we approach $\lim_{t\to\infty}p(t)$.

If the debtor has not defaulted by the year 2200, then it may be because some more favorable conditions may have set in than the present ones. Thus it is reasonable to think that $p(t)$ is decreasing for very large $t$. But a typical scenario where a concave down yield curve is expected is something like U.S. Treasury bonds, which have been solvent for much longer than the term $T$ (say, a couple of centuries compared to at most 30 years). So if the U.S. is solvent in 2042, the conditions are probably no better than they are now. It is more likely that the present stable conditions will have loosened a bit, forcing $p(t)$ to increase.