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I am really upset about an exercise, because i don't get it. Maybe someone can find my error?

We are talking about a single period model. Therefore our Asset is denoted by $S(0)=1$, $U=0.05$, $D=-0.05$, $R=0$ and $K=1$. Question seems easy: How much is the premium for the option $C(0)$?

I know: $C(0)=x_{C}S(0)+y_{C}A(0)$

where $x_{C} = \frac{S(0)(1+U)-K}{S(0)(U-D)}$, $y_{C}=\frac{(1+D)(S(0)(1+U)-K)}{A(0)(U-D)(1+R)}$

from solving the linear system $\begin{Vmatrix} xS(0)(1+U)+yA(0)(1+R)=S(0)(1+U)-K\\ xS(0)(1+D)+yA(0)(1+R)=0 \end{Vmatrix}$.

Provided solution should be $C(0)=0.025$, but i have no idea how, because by simply substituting given values the solution isn't correct. Are there any errors before?

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    $\begingroup$ If you accept my answer, could you pls click on the "tick mark" symbol next to my answer? That way the question will be marked as "answered". Thank you, $\endgroup$ Jun 7 '20 at 20:03
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The Stock price after the single period can be 1.05 or 0.95. If the stock ends up at 1.05, the option pay-off is 0.05, if the stock price ends up at 0.95 the option pay-off is zero. We want to figure out the price of the option by replicating it with the underlying Stock and a Bond (rates are zero, so the bond price at time zero is 1 and after the single period it is also 1). We wanna solve the following set of equations ($x$ is the number of stocks and $y$ is the number of bonds you hold to replicate the option pay-off at maturity):

$$x*0.95 +y = 0$$

$$x*1.05 + y = 0.05$$

Subtrack the first equation from the second and you get $x*0.1 = 0.05$, therefore $x=0.5$. You can now sustitute this into the second equation to get:

$$0.525 + y = 0.05$$

This solves to $y=-0.475$, therefore at maturity, if you are long 0.5 units of the Stock and short 0.475 units of the Bond, you replicate the option pay-off in both states.

Rates are zero so the option price at initial time is just 0.5 times the stock price - 0.475 * the bond price = 0.025. That's your answer.

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  • $\begingroup$ Fascinating. Nobody ever explained option pricing to me in such an approachable way. I'm sure there is way more to it, and the assumptions in this exercise are rather basic. But at least your explanation makes the whole thing accessible to me now. Thank you! $\endgroup$
    – FutForFut
    Jun 8 '20 at 6:32
  • $\begingroup$ @FutForFut: thanks. You'd be surprised how many people working in finance don't get the basic principle that an option price is the price of the replicating portfolio. People think it's the expectation of the pay-off at maturity, thinking it's some sort of probabilistic price, without realizing that the expectation is risk-neutral and rather than reflecting real-world probability, it actually computes the weights of the replicating portfolio. $\endgroup$ Jun 8 '20 at 7:06
  • $\begingroup$ Cool, thanks again. In the example, the stock price after the single period can be either 1.05 or 0.95. What if a large number of outcomes are possible? TSLA stock could be hundreds, if not thousands of different prices (at $0.01 resolution) a month from now? How would the calculation work in that case? $\endgroup$
    – FutForFut
    Jun 8 '20 at 7:19
  • $\begingroup$ @FutForFut: in that case, you start at the end of your binomial tree (the last period: i.e. end of the month for you, call it period n) and you work backwards towards period n-1 with only two outcomes. Say there is 10,000 possible outcomes at the period n in 0.01 "resolution". The top price on the tree is 1000, the second top one is 999.99. You treat these two branches as one period model and work backwards. You do the same with 999.99 & 999.98. When you solve all your equations, you know your option price all nodes n-1 & you repeat the process until you get to period zero. $\endgroup$ Jun 8 '20 at 7:26

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