Problem 32.45 Easy Free

Alternating Up and Down Moves

A stock starts at price $100\$100. It then falls 1%1\%, rises 1%1\%, falls 1%1\%, rises 1%1\%, and so on forever, alternating. What value does the price converge to?

Work it out, then compare with the answer.

SolutionShow the worked solution

Each fall multiplies the price by 0.990.99 and each rise by 1.011.01. A fall followed by a rise therefore multiplies it by

0.99×1.01=(1−0.01)(1+0.01)=1−0.012=0.9999<1.0.99\times 1.01=(1-0.01)(1+0.01)=1-0.01^2=0.9999<1 .

Let PkP_k be the price after kk moves. Then

P2n=100 (0.9999)n,P2n+1=99 (0.9999)n.P_{2n}=100\,(0.9999)^n,\qquad P_{2n+1}=99\,(0.9999)^n .

Both subsequences tend to 00 geometrically, since 0<0.9999<10<0.9999<1, so the whole sequence does too. The price never actually reaches zero, but it falls below any positive level eventually. For example, it halves after about ln⁡2/10−4≈6931\ln 2/10^{-4}\approx 6931 pairs of moves.

lim⁡k→∞Pk=0\boxed{\lim_{k\to\infty}P_k=0}

Key idea. Equal percentage moves up and down do not cancel. Their product is 1−r2<11-r^2<1, so alternating gains and losses of the same size steadily erode the value. This effect is called volatility drag.

Generalisation. With alternating moves of ±r\pm r for 0<r<10<r<1, the price after 2n2n moves is P0(1−r2)n→0P_0(1-r^2)^n\to 0.