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.

  • $\begingroup$ Thank you, so the discoubted stock price is the price of the stock At time 0 multiplied by the interest ? $\endgroup$ – user2505650 Apr 22 '15 at 7:24
  • $\begingroup$ No, the discounted stock price at $t$ is the quantity equal to $e^{-rt} S_t$. $\endgroup$ – Olorin Apr 22 '15 at 7:39
  • 1
    $\begingroup$ yes but what is its meaning ? what is the meaning of the discoubted price of an asset in general? $\endgroup$ – user2505650 Apr 22 '15 at 9:29
  • $\begingroup$ 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. $\endgroup$ – Olorin Apr 22 '15 at 11:27
  • $\begingroup$ 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". $\endgroup$ – user9403 May 22 '15 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.

  • $\begingroup$ Hi Gordon , let $0=t_0<t_1<\ldots<t_N=T$ . Do you use operator $E$ or $E_{t}$? $\endgroup$ – user16891 Jul 22 '15 at 19:43
  • $\begingroup$ @Farahvartish: Generally, we use $E$, but for conditional expectation with respect to information set $\mathcal{F}_t$, we short for $E_t$. $\endgroup$ – Gordon Jul 22 '15 at 20:33
  • $\begingroup$ I know it , Do you use change measure? $\frac{dQ}{dP}=\frac{S_T/B_T}{S_t/B_t}$? $\endgroup$ – user16891 Jul 22 '15 at 21:03
  • $\begingroup$ @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. $\endgroup$ – Gordon Jul 22 '15 at 21:06
  • $\begingroup$ you are right. I had not noticed it! $\endgroup$ – user16891 Jul 22 '15 at 21:18

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