# Financial Mathematics - Martingales example

Was hoping somebody could help me with the following question.

Prove that under the risk-neutral probability $\tilde{\mathsf P}$ the stock and the bank account have the same average rate of growth. In other words, if $S_0$ and $S_N$ are the initial and final stock prices, and $B_0$ and $B_N$ the initial and final bank prices, show that:

$$\tilde{\mathsf E}\left[\frac{S_N}{S_0}\right]=\tilde{\mathsf E}\left[\frac{B_N}{B_0}\right]=c$$ and find the constant c.

I have the following:

I know that the risk neutral (or non risk neutral) expectation of the bank account will simply be $B_N/B_0$, as the expectation of any bank related investment will simply be the same as whatever is in the bracket (there is no uncertainty in the bank).

Also, I know $B_N=B_0(1+r)^N$ ($B_N$ is equal to initial investment multiplied by interest rate to the power $N$), So I can simplify $$\tilde{\mathsf E}\left[\frac{B_N}{B_0}\right]= B_N/B_0 = \frac{B_0(1+r)^N}{B_0}=(1+r)^N.$$

My problem is trying to show that this is the case for the stock $\tilde{\mathsf E}\left[\frac{S_N}{S_0}\right] = (1+r)^N.$ As the stock is a martingale, I know I can say that:

$$S_0/(1+r)^0 = \text{(by multi step ahead property)} = \tilde{\mathsf E}\left[\frac{S_N}{S_0}\right].$$But I cannot work out what to do after this. I have found a way online that says this implies: $\tilde{\mathsf E}\left[\frac{S_N}{S_0}\right]=(1+r)^N$, but I cannot see how the previous statement implies this.

Would greatly appreciate any help.

• I'm not currently working on that model. The question does not specify the possible increments for a stock, so I imagine there's no set boundary.
– DFM
Mar 22, 2012 at 13:49
– SBF
Mar 22, 2012 at 13:53
• When you say "as the stock is a martingale" it is fully correct. It is the discounted stock process that is a (local) martingale under Q.
– Kolmo
Mar 22, 2012 at 14:02
• @Kolmo: so it is fully correct or not? :)
– SBF
Mar 22, 2012 at 14:06
• @Ilya of course it is not correct.
– Kolmo
Mar 22, 2012 at 14:09

The classic argument using risk-neutral pricing is to assume that discounted stock prices are $\tilde{P}$-martingales where $\tilde{P}$ is the risk-neutral probability measure.

Then, you know that

$$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$

by definition of a martingale process.

As the discounts are non-stochastic, you can safely remove it from the expectation, and as $S_t$ is $\mathcal{F_t}$-measurable, you can also include it freely in the expectation.

You then get

$$(1+r)^{T-t}=\tilde{E}[\frac{S_T}{S_t} | \mathcal{F}_t]$$

With your setup $T=N$ and $t=0$ and you get

$$(1+r)^{N}=\tilde{E}[\frac{S_N}{S_0} | \mathcal{F}_0]=\tilde{E}[\frac{S_N}{S_0}]$$

$$\widetilde{E}[\frac{S_{N}}{S_{0}}]=\widetilde{E}[\frac{S_{N}}{S_{N-1}}\frac{S_{N-1}}{S_{N-2}}...\frac{S_{1}}{S_{0}}]=\widetilde{E}[\frac{S_{N}}{S_{N-1}}]\widetilde{E}[\frac{S_{N-1}}{S_{N-2}}]...\widetilde{E}[\frac{S_{1}}{S_{0}}]=(1+r)(1+r)...(1+r)=(1+r)^{N}$$

• To understand that you need to know that under the risk-neutral measure, the expected return on an asset price is the risk-free rate $r$, by definition.
– SRKX
Apr 5, 2012 at 8:10