Risk-Neutral Martingales and Expected Stock Growth
Summary
Under a risk-neutral probability measure, discounted stock prices are martingales. With a deterministic bank account growing at the per-period risk-free rate, the conditional expectation of the stock’s future discounted value equals its current value. Rearranging this relation gives the expected future stock-price ratio, conditional on current information, as the bank account’s growth factor over the same horizon.
At the initial date, this implies that the expected stock growth over the full period equals the risk-free accumulation factor, matching the bank account’s average growth. The answer derives the result by removing deterministic discount factors from the expectation and using the fact that the current stock price is already known. This conclusion relies on the risk-neutral measure, a deterministic rate, and the martingale pricing assumption; it does not say that realized stock returns are certain or equal to the risk-free return.
Key ideas
- Discounted stock prices are martingales under the risk-neutral measure.
- With deterministic interest rates, discount factors can be taken outside expectations.
- The conditional expected stock growth over a horizon equals the bank account growth factor over that horizon.
- The result concerns expected returns under risk-neutral pricing, not realized stock performance.
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# Financial Mathematics - Martingales example
# 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.
## Answer by SRKX (score 8)
https://quant.stackexchange.com/a/3180
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
$$(1+r)^{N}=\tilde{E}[\frac{S_N}{S_0} | \mathcal{F}_0]=\tilde{E}[\frac{S_N}{S_0}]$$
## Answer by Kumar (score 0)
https://quant.stackexchange.com/a/3186
$$ \widetilde{E}[\frac{S_{N}}{S_{0}}]=\widetilde{E}[\frac{S_{N}}{S_{N-1}}\frac{S_{N-1}}{S_{N-2}}...\frac{S_{1}}{S_{0}}]=\widetilde{E}[\frac{S_{N}}{S_{N-1}}]\widetilde{E}[\frac{S_{N-1}}{S_{N-2}}]...\widetilde{E}[\frac{S_{1}}{S_{0}}]=(1+r)(1+r)...(1+r)=(1+r)^{N} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.