Finding the expected time for a stock to go from $\$25$ to $\$18$ given there is a support level at $20 with upward and downward biases.

174 Views Asked by At

This problem is adapted from Stochastic Calculus and Financial Applicationsby J. Michael Steele, Springer, New York, 2001, Chapter 1, Section 1.6, page 9.

Consider a naive model for a stock that has a support level of \$20/share because of a corporate buy-back program. Suppose also that the stock price moves randomly with a downward bias when the price is above \$20, and randomly with an upward bias when the price is below \$20. To make the problem concrete, we let $Y_n$ denote the stock price at time $n$, and we express our stock support hypothesis by the assumptions that

\begin{eqnarray*} \Pr[ Y_{n+1} = 21 | Y_{n} = 20] &=& 9/10 \\ \Pr[ Y_{n+1} = 19 | Y_{n} = 20] &=& 1/10 \end{eqnarray*}

We then reflect the downward bias at price levels above \$20 by requiring that for $k > 20$:

\begin{eqnarray*} \Pr[ Y_{n+1} = k+1 | Y_{n} = k ] &=& 1/3 \\ \Pr[ Y_{n+1} = k-1 | Y_{n} = k ] &=& 2/3. \end{eqnarray*}

We then reflect the upward bias at price levels below \$20 by requiring that for $k < 20$:

\begin{eqnarray*} \Pr[ Y_{n+1} = k+1 | Y_{n} = k ] &=& 2/3 \\ \Pr[ Y_{n+1} = k-1 | Y_{n} = k ] &=& 1/3 \end{eqnarray*}

I would like to calculate the expected time, $T_{25,18}$ for the stock to fall from $\$25$ through the support level all the way down to $\$18$.

My first step is to use first-step analysis (no pun intended). This will give me a recursive set of 9 equations. However, one of the hints given is to show that the expected time to go from $\$25$ to $\$20$ is $T_{25,20} = 15$ steps and that $T_{21,20} = 3$. Is it claimed that these are trivial to find. However, I am really not sure how to do this part. There appears to be no upper boundary above $\$25$ -- Does anyone have any hints as to how I can find this? Thanks.

1

There are 1 best solutions below

1
On BEST ANSWER

Let $\tau=\inf\{n>0: Y_{n+1}<Y_n\mid Y_0=25\}$. Then $$\mathbb E[\tau] = 1 + \frac23 + \frac13\mathbb E[\tau+1]\implies \mathbb E[\tau] =3.$$ It follows that $T_{21,20}=\tau=3$ and $T_{25,20}=5\tau=15$. The quantities $T_{20,19}$ and $T_{19,18}$ may be computed by a similar argument, and $$T_{25,18} = T_{25,20}+T_{20,19}+T_{19,18}. $$