Probability of market movement / Trends

Question

If you imagine a scale from -100 to 100, if the market has moved up from 0 to 40, what is the probability is will continue to 100?

There is a 50-50 chance to move up or down from 0, but what is the probability of moving to 100 when it has already moved to 40?

This may be a simple question, but I am struggling! Can someone share the the algorithm that solves this problem?

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Is the answer 70%, ie. if at 0 it is 50-50 to go up or down, if it moves to 40 there is 70% chance it will rise to 100 and a 30% chance it will not?

Answer 1

It's a well established 'empirical fact' that there is no significant autocorrelation in market returns. That is, knowing that the market moved up in the last period provides no information about what it will do in the next period.

So the answer to your question (as backed up by a lot of data) is that it is still 50/50 whether the market goes up or down next, no matter what just happened.

You haven't really asked a mathematical question, which is why I haven't give you a mathematical answer.

Answer 2

If (and this is a big if) you are considering a symmetric $\pm1$ random walk or a driftless Brownian motion starting from $x=40$ and you are wondering about the probability $u(x)$ that it hits $x_1=100$ before hitting $x_0=-100$, then indeed the answer is $$ u(x)=\frac{x-x_0}{x_1-x_0}=70\%. $$ Briefly, an elementary method in the random walk case is to compute $u(x)$ for every integer starting point $x$ between $x_0$ and $x_1$. For every such $x$, one has $50\%$ chances that the first step is to $x+1$ and $50\%$ chances that the first step is to $x-1$, and from there, one looks for the probability $u(x\pm1)$ to hit $x_1$ before $x_0$. Hence, $u(x)=\frac12(u(x+1)+u(x-1))$.

This means that $x\mapsto u(x)$ is a straight line on the integer interval $[x_0,x_1]$. Obviously, $u(x_0)=0$ and $u(x_1)=1$ hence $u(x)$ is as written above.

A slightly more advanced method is to consider the position $X_n$ of the random walk at time $n\wedge\tau_0\wedge\tau_1$ where $\tau_0$ is the first hitting time of $x_0$ and $\tau_1$ is the first hitting time of $x_1$. This means that the random walk performs equiprobable independent $\pm1$ steps and that one stops it as soon as it hits $x_0$ or $x_1$. Then $(X_n)$ is a bounded martingale hence its expectation does not depend on $n$. Now $X_0=x$, $X_n\to x_1$ on $S=[\tau_1<\tau_0]$, $X_n\to x_0$ on $[\tau_0<\tau_1]=S^c$, and one gets $$ x=X_0=\lim\limits_{n\to\infty}E(X_n)=E(\lim\limits_{n\to\infty}X_n)=x_0P(S^c)+x_1P(S). $$ Since $P(S)+P(S^c)=1$, this yields the result.

In the Brownian motion case, one can still use the idea of the second method, replacing finite differences by a differential operator: one gets that $u(x_0)=0$, $u(x_1)=1$ and $u''(x)=0$ for every real number $x$ in the interval $(x_0,x_1)$. This means that $u$ is affine in this case as well hence $u(x)$ is given by our first displayed formula, where now the argument $x$ is a real number.