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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.

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  • $\begingroup$ 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. $\endgroup$
    – DFM
    Mar 22, 2012 at 13:49
  • $\begingroup$ please put the definition of the martingale measure, that should help you $\endgroup$
    – Ilya
    Mar 22, 2012 at 13:53
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    $\begingroup$ 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. $\endgroup$
    – Kolmo
    Mar 22, 2012 at 14:02
  • $\begingroup$ @Kolmo: so it is fully correct or not? :) $\endgroup$
    – Ilya
    Mar 22, 2012 at 14:06
  • $\begingroup$ @Ilya of course it is not correct. $\endgroup$
    – Kolmo
    Mar 22, 2012 at 14:09

2 Answers 2

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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

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

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

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  • $\begingroup$ 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. $\endgroup$
    – SRKX
    Apr 5, 2012 at 8:10

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