0
$\begingroup$

This question comes from a Financial Economics exam and I'm very confused about a state-price deflator which doesn't seem to exist. I've included the whole question for completeness, but my actual question is quite concise.

An investment bank has issued a two-month European call option on a non-dividend-paying share (whose current price $S_0 = 100$) with a strike price $K = 98$. The risk-free rate of interest $r = 8\%$ per annum and the expected return on the share is $15\%$ per annum, both continously compounded. The share price process is assumed to follow Geometric Brownian motion with the annual standard deviation of the change in the log share price equal to $25\%$.

tldr: $S_0 = 100$, $K = 98$, $r = 8\%$, $\sigma = 0.25$ and $dS_t = S_t(0.15dt + 0.25dW_t)$

(i) Construct a two period recombining binomial model where each period is one month, and hence derive the state price deflators at time $t = 2$. The parameters determining the share price after an up-jump and down-jump should be those which calibrate the model to the above standard deviation.

(ii) Use the state price deflators from (i) above to derive the value of the option at time $t = 0$.

(iii) Confirm that the same option value is obtained by means of risk-neutral valuation

Now, I can price this option without using state price deflators. A summary of my workings so far is:

Using the Cox-Ross-Rubinstein assumption,, the size of the up-jump: $$u = e^{\sigma\sqrt{\delta t}} = e^{0.25/\sqrt{12}} = 1.07484 = 1/d$$ Risk-neutral probability of an up-jump: $q = \frac{e^{r\delta t} - d}{u - d} = 0.52826$

From here it is easy to work out the risk-neutral probabilities for each of the states: $$\mathbb{Q}(\{uu\}) = 0.27907$$ $$\mathbb{Q}(\{ud\}) = \mathbb{Q}(\{du\}) = 0.24920$$ $$\mathbb{Q}(\{dd\}) = 0.22253$$

From there, risk-neutral pricing is easy.

However, what are the state-price deflators in this example?

From my notes: the state price deflator in the Binomial model is:

$$A_n = e^{-rn}(\frac{q}{p})^{N_n}(\frac{1-q}{1-p})^{n-N_n}$$

where $N_n$ is the number of upsteps up to time $n$ and $p$ is the real-world probability of an up-step.

However, because we are given that $S_t$ has GBM dynamics in the real-world probability measure: $$p = \mathbb{P}[\text{up-step}] = \mathbb{P}[S_{t+1} = u \cdot S_t] = 0$$

My question: How do we define the state-price deflator when the assumed real-world distribution is continuous but we're working in the (necessarily discrete) Binomial model?

$\endgroup$
2
$\begingroup$

first you have to find the p, u and d that match standard deviations in part (I). The problem is a little underdefined in that even if you match mean and standard deviation precisely, there are multiple solutions. Impose an extra condition to get all three.

Once this is done compute $q$ via $$ q = \frac{e^{r \delta t} - d}{u-d}. $$ The state-price deflator is then found from the ratio $q/p.$

$\endgroup$
  • $\begingroup$ But what is $p$? $\endgroup$ – Stanley Jun 1 '15 at 22:39
  • $\begingroup$ that is why I said the problem was underdefined. You need to impose an extra condition to make the solution for u d and p unique. eg match 3 moments, take p=0.5, take u = 1/d. $\endgroup$ – Mark Joshi Jun 1 '15 at 23:42
0
$\begingroup$

I guess for you to obtain the value for the real world probability measure P, the expected rate of return should be given...else the value of P should be given.

$\endgroup$
0
$\begingroup$

When using, for example, a GBM model, the stochastic process for deflator is:

$$\mathrm{d}Dt = Dt\cdot \left[-r\mathrm{dt}-(\frac{m-r}{s}) \mathrm{d}Zt \right]$$

where $r$ denotes the risk-free rate, $m$ and $s$ are the drift and vol respectively.

If $r$, $m$ or $s$ are time-dependent or even stochastic, use the relevant value for the given projection period or simulation. In a multivariate case, this generalizes to a matrix form, where instead of scalar $m$ you get a vectors of m's, instead of $r$ you get a vector of the same value $r$, and instead of dividing by $s$ you multiply by the cholesky of the inverse covariance matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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