I got a question and corresponding solution, but have some difficulties in understand the lognormal distribution part of it, so I really appreciate your advice:

Question: assume zero interest rate and a stock with current price at 1 dollar that pays no dividend. When the price hits level H(H>0) for the first time you can exercise the option and received 1 dollar. What is this option worth to you today?

Solution: since the stock price follows a geometric Brownian motion under risk-neutral measure $dS = rSdt+σSdW(t)$. Since r=0, $dS = σSdW(t)$, so $d(lnS)=-0.5*σ^2dt+σdW(t)$ When t=0, we have $S_0=1,ln(S_0)=0$. Notice that S is a martingale under the risk-neutral measure, but $lnS$ has a negative drift.

Here is my first doubt: if I understand correctly, the reason why $dlnS$ has a negative drift while $dS$ does not have is because: dlnS is continuously compounded rate of stock price, due to continuously compounded feature, it takes into account volatility (or standard deviation), so its real drift should be subtracted by this volatility component, I'm wondering if my understanding is correct?

(continued from the solution) The reason is that $lnS$ follows a normal distribution, but $S$ itself follows a lognormal distribution, which is positively skewed. As $T$ approaches positive infinity, although the expected value of $S_T$ is 1, the probability that $S_T>=1$ actually approaches 0. Here is my second doubt: how do we know the probability that $S_T>=1$ actually approaches 0?


The drift of $\mathrm{d}\ln(S_t)$ is indeed $r-\frac{1}{2}\sigma^2$ which is always negative if $r=0$. The extra $-\frac{1}{2}\sigma^2$ has many explanations. You could see it as a convexity correction (see Jensen's inequality) or martingale correction. Without it, $(S_t)$ wouldn't be a martingale.

For the second part, note that $\ln(S_T)\sim N\left(\ln(S_0)-\frac{1}{2}\sigma^2T,\sigma^2 T\right)$. Then, for $Z\sim N(0,1)$, \begin{align*} \mathbb{Q}\left[ \{S_T\geq1\}\right] &= \mathbb{Q}\left[ \{\ln(S_T)\geq0\}\right] \\ &= \mathbb{Q}\left[ \left\{\ln(S_0)-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}Z\geq0\right\}\right] \\ &= \mathbb{Q}\left[ \left\{Z\geq \frac{-\ln(S_0)+\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}\right\}\right] \\ &= 1- \Phi\left(-\frac{\ln(S_0)-\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}\right) \\ &= \Phi\left(\frac{\ln(S_0)-\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}\right). \end{align*} You should note that this is the same term as $\Phi(d_2)$ in the Black Scholes formula for $K=1$ and $r=0$. Thus, \begin{align*} \lim_{T\to\infty} \mathbb{Q}[\{S_T\geq1\}] = 0. \end{align*} This makes sense. Since the returns have a negative drift, you expect the stock price to decline over time. Hence, the probability of $S_T$ being greater than any positive constant $\varepsilon>0$ tends to zero as $T\to\infty$.

  • $\begingroup$ I think there is a typo in step 4 you have the minus wrong? $\endgroup$ – emcor Dec 16 '19 at 19:34
  • $\begingroup$ @emcor Where exactly? I used $\mathbb{Q}[\{Z\geq x\}]=1-\mathbb{Q}[\{Z\leq x\}]=1-\Phi(x) = \Phi(-x)$ or, if your prefer, $\mathbb{Q}[\{Z\geq x\}] = \mathbb{Q}[\{-Z\leq -x\}] = \mathbb{Q}[\{Z\leq -x\}]=\Phi(-x)$ for $x=\frac{-\ln(S_0)+\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}$. $\endgroup$ – Kevin Dec 16 '19 at 19:42
  • $\begingroup$ sorry my mistake, but it is a bit misleading that you are unnecessarily changing the sign twice in $1- \Phi\left(-\frac{\ln(S_0)-\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}\right) \\$ $\endgroup$ – emcor Dec 17 '19 at 11:28

A quick way is to use the following property of the B.M: $\lim_{t \rightarrow \infty} \frac{W_t}{t} = 0$ a.s.

The logarithm of the solution of the Black-Scholes SDE (when $r=0$) is $\ln S_t = - \frac{\sigma^2}{2}t + \sigma W_t.$ Hence, $$\lim_{t \rightarrow \infty} \frac{\ln S_t}{t} = \lim_{t \rightarrow \infty} - \frac{\sigma^2}{2} \frac{t}{t} + \lim_{t \rightarrow \infty} \frac{\sigma W_t}{t} = - \frac{\sigma^2}{2}. \text{ a.s. }$$ Thus, $\lim_{t \rightarrow \infty} S_t = \lim_{t \rightarrow \infty} e^{- \frac{\sigma^2}{2}t} = 0$ a.s., which implies the required result.

Edit based on comments:

We have the stochastic process $\{S_t(\omega) \}_{t \geq 0}$ on a probability space $(\Omega, \mathscr{F}, \{ \mathscr{F}_t^S \}_{t \geq 0}, P)$ where $\{ \mathscr{F}_t^S \}_{t \geq 0}$ is the filtration generated by $\{S_t(\omega)\}_{t \geq 0}$ and $\mathscr{F} = \sigma(\cap_{t \geq 0}\mathscr{F}_t^S).$ For every $t,$ $S_t(\omega)$ is is a random variable. When we write in short $\{ S_t = 0 \},$ we mean $\{\omega \in \Omega: S_t(\omega)=0 \}.$ Similarly, $P( S_t = 0)$ is the short way of writing $P(\{ \omega \in \Omega: S_t(\omega)=0 \ \}).$ The set $\{\omega \in \Omega: S_t(\omega)=0 \}$ belongs to $\ \mathscr{F}_t \subset \mathscr{F}$ and it has associated a probability (since $P:\mathscr{F} \rightarrow [0,1]$). By definition (see paragraph 'Formal definition' here https://en.wikipedia.org/wiki/Almost_surely )

An event $E \in \mathscr{F}$ happens almost surely if $P(E)=1.$

In our case the event is $\{\omega \in \Omega: \lim_{t \rightarrow \infty} S_t(\omega) = 0 \}.$ Hence, to say $\lim_{t \rightarrow \infty} S_t = 0$ .a.s. is exactly the same as $P(\{\lim_{t \rightarrow \infty} S_t = 0\})=1$. Then obviously we have that $P(\{\lim_{t \rightarrow \infty} S_t \geq 1 \})=0.$ In my opinion, this is what the exercise asks. However I think the other answer is perfectly correct because the cumulative distribution function of a log-normal r.v. is continuous and we can "pass the limit inside the probability function".

  • $\begingroup$ Can you please comment on the probability aswell? How does the limit of $S_t$ imply a $0$ probability? $\endgroup$ – emcor Dec 17 '19 at 11:33
  • $\begingroup$ By definition of convergence almost surely, $\lim_{t \rightarrow \infty} S_t = 0$ a.s. is equivalent to $P(\lim_{t \rightarrow \infty} S_t = 0)=1$, so $P(\lim_{t \rightarrow \infty} S_t \geq 1) =0.$ $\endgroup$ – user39119 Dec 17 '19 at 12:35
  • $\begingroup$ The question was w.r.t. $\lim P(\cdot)$ not $P(\lim \cdot)$, I am not sure this is fully interchangeable? $\endgroup$ – emcor Dec 17 '19 at 13:39
  • $\begingroup$ @UBM,thanks a lot for the detailed explanation, if you don't mind, could you please show us why $\lim_{t \rightarrow \infty} S_t = \lim_{t \rightarrow \infty} e^{- \frac{\sigma^2}{2}t}$ holds? $\endgroup$ – M00000001 Dec 17 '19 at 14:40
  • $\begingroup$ @emcor: I disagree. When $T$ approaches positive infinity, $\lim_{T \rightarrow \infty} S_T$ is a random variable. So in my opinion they are asking for the probability of the event $\{\omega \in \Omega: \lim_{T \rightarrow \infty} S_T(\omega) \geq 1\}.$ Anyway, the cumulative distribution function of a normal random variable is continuous so 'limit of the probability' is equal to 'probability of the limit'. $\endgroup$ – user39119 Dec 17 '19 at 15:34

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