# The right choice when the price of a stock follows a random walk

I've got the following question:

Suppose the price of a stock either rises or falls by the same percentage for each day. Suppose there is no dividend and the interest rate is 0. Should I buy the stock now or sell it? Is there no difference?

I think there were more conditions to this problem but I cannot remember. Owning the stock gives some rights of the company, so I think we should also assume that the prospect of the company is neutral as well... I couldn't understand the intention and the answer of the problem. I would really appreciate if someone can explain it.

• I am voting to close this question for being too basic. Hint: Let $x$ be the percentage in question. Is $(1 + x) / (1 - x)$ greater or smaller than one? – LocalVolatility Nov 1 '17 at 7:56
• Don't you mean $(1+x) \times (1 -x)$ @LocalVolatility? – Bob Jansen Nov 1 '17 at 8:01
• Yes - absolutely. Thanks for catching it! – LocalVolatility Nov 1 '17 at 8:02
• Can you clarify, for $0 < x < 1$, $(1+x) \times (1-x) > (1+x)^2 \times (1-x)^2$, right? – Bob Jansen Nov 1 '17 at 10:24

It makes no difference. Starting with a capital of 1, let $X_i$ be the multiplying factor for the $i$th day, so $X_i\in\{1+r,1-r\}$ with each possibility having probability 1/2. The expected capital after one day is $$\mathbb E(X_1)=\frac12((1+r)+(1-r))=1.$$ After $n$ days, your capital is $X_1X_2\cdots X_n$, and $$\mathbb E(X_1\cdots X_n)=\mathbb E(X_1)\cdots\mathbb E(X_n)=1$$ since the days are independent.

### Clarifying notes

• You might think,

"$(1+r)(1-r)=1-r^2<1$ so as the stock goes up and down, I lose money"

but note that if the stock goes down, then down again, you have $$(1-r)^2=1-2r+r^2>1-2r,$$ so you get more than what you'd get using a "simple interest" idea, and this together with the up/up case cancels out the loss in the up/down case.

Indeed, for $n=2$ the expected capital is $$\frac14[(1+r)^2+2(1+r)(1-r)+(1-r)^2)]=1.$$

• Of course if you like or dislike risk then you may want to buy or not buy, respectively, but the expected capital remains 1.

• On the other hand if the multipliers were not $\{1+r,1-r\}$ but $\{1+r, \frac1{1+r}\}$ then it would make sense to buy, as the expectation after one day would be $$\frac12\left(1+r+\frac1{1+r}\right)=\frac12\frac{(1+r)^2+1}{1+r}=1+\frac{r^2}{2(1+r)}>1.$$

• I already, like your answer better. However, for $n=2$, three scenario's have a negative return and one a positive. In an interview situation I would definitely say something about how that would make me feel, risk-wise. – Bob Jansen Nov 1 '17 at 16:15
• @BobJansen that's a good point for the one-time, short-term investor. – Bjørn Kjos-Hanssen Nov 1 '17 at 16:42