Risk-Neutral Expected Cash Flows and Derivative Pricing

# Why is it enough to know the expected present value of cash flow in risk-neutral framework to price derivatives?


# Why is it enough to know the expected present value of cash flow in risk-neutral framework to price derivatives?












Wilmott book states that its enough to know the expected present value of all cash flow in risk-neutral framework to price derivatives.

As I know, to obtain arbitrage-free market we need our discounted price process to be martingale under the risk neutral $Q$ measure. Why does that imply the statement?

## Answer by Probilitator (score 2, accepted)

https://quant.stackexchange.com/a/11462

To give you another perspective:

Let us assume that the world had only one risky/noisy asset $S(t)$ and let us further assume that at time $T$ our process cann only have $n$ states - namely $(S_1, \dots, S_n)$ and that the interest rate was flat and given by $r$

Now let's say we have a payoff funtion $f(x): \mathbb{R}\to\mathbb{R}$.

Working under the risk neutral measure $Q$ the time $t$ price of the derivative paying $f(S(T))$ at time $t=0$ is given by $$ V(0)=e^{-rT}\mathbb{E}^Q[f(S(T))] $$

Now we know that $S(T)$ only has $n$ different states and can thus decompose above expectation into $$ V(0)=e^{-rT}\mathbb{E}^Q[f(S(T))]=e^{-rT}\sum_{i=1} \mathbb{E}^Q[f(S_i)] $$

> Thus our price is determined by the expected present values of the different cash-flows that can be generated by our instrument/product.

In above case one would actually already know the price for every function $g(x):\to\mathbb{R}\to\mathbb{R}$ if all the probabilities $P_i=\mathbb{P}^Q(S(T)=S_i)$ were known.

The price would then be given by $$ V(0)=e^{-rT}\mathbb{E}^Q[g(S(T))]=e^{-rT}\sum_{i=1} P_ig(S_i) $$

## Answer by pincopallino (score 2)

https://quant.stackexchange.com/a/11461

This tackles the second part of your question:

In a world were interest rates are always zero (for simplicity sake), if the discount price process is a martingale, we have:

$E[X_T | F_t] = X_t$

In an arbitrage free world, every price process is a martingale in the risk-neutral measure. Having martingale price processes means that if we build a hedged portfolio of value $P=0$ today, for the expected value to be possibly positive and the expected value to be zero, there must be positive probability of the portfolio ending up having a loss. This rules out the existence of an arbitrage by its definition (see definition here): generating profits without the risk of a loss.

Regarding the first question: This is a financial engineering statement. The modus-operandi is to decompose financial product in their cash-flows and then discount them to present value. Not sure if I could add much more on that.

## Answer by berkorbay (score 0)

https://quant.stackexchange.com/a/11466

Simply put, it is all about modelling the underlying asset(s) and option price is the discounted value of that asset's price in the future. Martingale measure pricing is somewhat close to "steady-state" from markov chain literature. If you know the expected steady-state price of your asset in the future you can discount it by the risk-free rate to get your fair price.