# Discounted Stock Price

I have the following Question :

Prove that under the risk-neutral probability p the stock and the banjaccount have the same average rate of growth. In other words, if $S_0 , S_N$ are the initial and final stock prices and $B_0 , B_N$ the initial and final bank prices , show that :

$$E[S_N / S_0 ] = E[B_N / B_0 ] = c$$

Hint : The discounted stock price is a martingale under P.

Could you explain to me what is the discounted stock price ?

Hope you will not mind if I place myself in continuous time. The discounted stock price at $T$ is $e^{-rT}S_T$. As you know that it is a martingale, you have that $\mathbf{E}^{\mathbf{P}}[e^{-rT}S_T | \mathscr{F}_t] = e^{-rt} S_t$ when $t\leq T$ which you can rewrite as $\mathbf{E}^{\mathbf{P}}\left[\frac{e^{-rT}S_T}{e^{-rt} S_t} | \mathscr{F}_t\right] = 1$ or $\mathbf{E}^{\mathbf{P}}\left[\frac{S_T}{S_t} | \mathscr{F}_t\right] = e^{r(T-t)}$ and $e^{r(T-t)}$ is but $\mathbf{E}^{\mathbf{P}}\left[\frac{B_T}{B_t} | \mathscr{F}_t\right]$. Finally, taking $t=0$ gives you the equality you are looking for.

• Thank you, so the discoubted stock price is the price of the stock At time 0 multiplied by the interest ? Commented Apr 22, 2015 at 7:24
• No, the discounted stock price at $t$ is the quantity equal to $e^{-rt} S_t$. Commented Apr 22, 2015 at 7:39
• yes but what is its meaning ? what is the meaning of the discoubted price of an asset in general? Commented Apr 22, 2015 at 9:29
• Take a deterministic price $F$ at time $t$. It's time $0$ value is $e^{-rt} F$ and it is the value of cash that you have to have at time $0$ so that, letting grow this cash on the bank account up t o $t$, you will get $F$. This work because $F$ is deterministic. So that for $S_t$ it doesn't work, as $S_t$ is a random variable. But, by abuse, $e^{-rt} S_t$ is nevertheless called disounted asset price. Commented Apr 22, 2015 at 11:27
• A "discounted asset" is simply an asset denominated in another asset (this is called a numeraire). In this case, the terminal payout of the "Bank Price" is a dollar which is denominated by the growth of the "Bank Price". The stock is likewise denominated by the "Bank Price". By no arbitrage, any asset denominated by another asset is a martingale under the measure induced by the denominating asset. In this case, the common name for the measure induced by the "Bank Price" is termed the "risk neutral measure". Commented May 22, 2015 at 10:19

Let $S_t$ and $B_t$ be respectively the stock price and the money market account value at time $t$. Then $S_t/B_t$ is called the discounted stock price. Note that \begin{align*} E\left(\frac{S_N}{S_0}\right) &= E\left(\frac{S_N}{B_N} \frac{B_N}{B_0}\right)\frac{B_0}{S_0}\\ &= E\left(\frac{S_N}{B_N}\right) E\left(\frac{B_N}{B_0}\right)\frac{B_0}{S_0} + Cov\left(\frac{S_N}{B_N}, \frac{B_N}{B_0}\right)\frac{B_0}{S_0}, \end{align*} where $Cov(\,)$ is the covariance operator. If the interest rate is deterministic, then \begin{align*} E\left(\frac{S_N}{S_0}\right) &= E\left(\frac{S_N}{B_N}\right) E\left(\frac{B_N}{B_0}\right)\frac{B_0}{S_0}\\ &= E\left(\frac{B_N}{B_0}\right). \end{align*} However, if the interest rate is stochastic, this conclusion may not be true.

• Hi Gordon , let $0=t_0<t_1<\ldots<t_N=T$ . Do you use operator $E$ or $E_{t}$?
– user16891
Commented Jul 22, 2015 at 19:43
• @Farahvartish: Generally, we use $E$, but for conditional expectation with respect to information set $\mathcal{F}_t$, we short for $E_t$. Commented Jul 22, 2015 at 20:33
• I know it , Do you use change measure? $\frac{dQ}{dP}=\frac{S_T/B_T}{S_t/B_t}$?
– user16891
Commented Jul 22, 2015 at 21:03
• @Farahvartish: There is no measure change involved here. Why do you ask this? The poster ask to show that $E[S_N / S_0 ] = E[B_N / B_0 ] = c.$ What I said is that this is only true with certain conditions, for example, deterministic interest rates. Commented Jul 22, 2015 at 21:06
• you are right. I had not noticed it!
– user16891
Commented Jul 22, 2015 at 21:18