1
$\begingroup$

The following question, which is not homework, was taken from a past paper for a module I will soon be sitting:

Consider a Black-Merton-Scholes stochastic market with drift $\mu = 1$, volatility $\sigma^2 = 1$ and interest rate $r = 0$. On this market there is a European option with a certain expiry $T > 0$ and a payoff $$ \sqrt{1 + \left( \frac{T}{2} + \ln S_T \right)^2} $$ where S stands for the stock price. Assuming $S_0 = 1$, show that the price of this option may not exceed $\sqrt{1 + T}$.

I am struggling to answer this question. My attempt so far is as follows:

Firstly, since we are considering the stock price under the BMS market model, I believe that $S_t$ will be given by $$ S_t = S_0 \exp \left( \left( r - \frac{\sigma^2}{2} \right)t + \sigma W_t \right) = \exp \left( - \frac{t}{2} + W_t \right) $$ Thus, denoting by $V_t$ the value of this option at time $t$, we have $$ V_t = \sqrt{1 + \left( \frac{t}{2} + \ln S_t \right)^2} = \sqrt{1 + \left( \frac{t}{2} + \left( - \frac{t}{2} + W_t \right) \right)^2} = \sqrt{1 + W_t^2} $$ and the price of this option at $t=0$ would thus be given by $$ V_0 = e^{-rT} \mathbb{E} [V_T] = \mathbb{E} \left[ \sqrt{1 + W_T^2} \right] $$

Assuming that I am correct thus far, I am now required to show that $$ \mathbb{E} \left[ \sqrt{1 + W_T^2} \right] \leq \sqrt{1+T} $$ How might I do this?

$\endgroup$
3
  • $\begingroup$ BSM not BMS. And it is not called a market model. $\endgroup$
    – Dom
    May 24, 2018 at 5:03
  • $\begingroup$ In my lecture notes it is Black-Merton-Scholes (BMS). $\endgroup$
    – M Smith
    May 24, 2018 at 8:42
  • $\begingroup$ Your lecturer is wrong. $\endgroup$
    – Dom
    May 24, 2018 at 16:43

1 Answer 1

3
$\begingroup$

You have a typo in the last two lines: $W_t$ should be $W_T$. Your approach is correct so far.

Use Jensen for the last step: $E[\sqrt{1+W_T^2})]^2\le E[1+W_T^2]=1+T$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.