Some Notes On Mathematical Finance

Intro¶

hopefully after reading this you will be able to value “fincial produucts” so for an insurance comapny this may be things like a IGR / and for banking these may be derivatives or other such complex products

The big Idea¶

How do you price a product that does not exist in the market How do I hedge such a an instrument?

e.g cos(tesla sahre) how do I find the price of this

A motivating example¶

consider a world with oonly two time points

Undefined control sequence: \hdots at position 19: …gin{align}
t=0 \̲h̲d̲o̲t̲s̲ ̲t=T
\end{align}

\begin{align}
t=0 \hdots t=T
\end{align}

where t=0 is today and some random point in the future t=T

we will also aumme that ther are only two possible “states” at time t=T, a “good” state and a “bad” state

\begin{align} \mathbb{P}(g)&=0.7 \mathbb{P}(b)&=0.3 \end{align}

(2)

we assume a “primary/trivial assets” of a share and a bank account

\begin{align} S\longrightarrow \text{share} B\longrightarrow \text{bank account} \end{align}

(3)

initiall the share is 100 bucks

\begin{align} S_{0} = 100 \end{align}

(4)

and the future value of the share could be one of 2 possibilities

\begin{align} S_{1} = 140 if g =90 if b \end{align}

(5)

and the bank account earns 6% mathmatically we can denote this by

\begin{align} B_{0} = 1 \end{align}

(6)

and

\begin{align} B_{T} = 1.06 \end{align}

(7)

all together we can represent this in a tree diagram like so

\usepackage{forest} \begin{forest} for tree={ draw, rounded corners, align=center, edge={-stealth, thick}, l sep=1cm, % level separation s sep=0.5cm % sibling separation } [Root Node [Child 1] [Child 2 [Grandchild 1] [Grandchild 2] ] [Child 3] ] \end{forest}

from this we can now answer some questions what is the expected return of the bank account? this is obviously 6%

a harder one woul be what is the expected return of the share

Undefined control sequence: \plus at position 57: …  40\times 0.7 \̲p̲l̲u̲s̲ ̲(-10) \times 0.…

\begin{align}
expected value of share  &=  40\times 0.7 \plus (-10) \times 0.3 \\
                          &=  25
\end{align}

therfore

\begin{align} expected return of share &= 25/100 \\ &= 25% \end{align}

(9)

now let us consider another non primary asset, a derivative, i.e. a complicate instrument bbased on primary assets in this scenario we will introduce a call option with a strike price of {math}K=110

So if the future the price of the share is 2000 then we would obviously excercise our right to but it at 110. Conversely if the price of the share in the future is 20 then we wwould not excercise the right to buy it at 110 because we can just go to the open market and buy it for 20

mathematically we can express this terminal payoof condition as

\begin{align} C_{T} &= \max{S_{T}-110,0} &= S_{T} -K if S_{T}\geq K or 0 otherwise \end{align}

(10)

so in the simplistic two state example we have described above

\begin{align} C_{T} = 30 if good \end{align}

(11)

\begin{align} C_{T} = 0 if bad \end{align}

(12)

our goal in life is to answer the question

what is {math} C_0?
How do you hedge {math}C?

Let us try to find C_0 using expected present value

\begin{align} C_{0} &= 30 \times 1.06^{-1} \times 0.7 + 0 \times 1.066{-1} \times 0.3 &= 19.81 \end{align}

(13)

this seems reasonable however using expected present value on the sahre i.e. discounting back S_{T} back to find S_0 does not give you S_0 = 100 hence if EPV does not hold for the share we cannot expect it to hold for the Call

even worse, the probabilities we used were completely made up out of thin air. these are the real world possibilities and hence any person you ask what the proability of the good state would be would give you different answers (i.e. some might say 0.7 like us but others would say 0.65 or 0.95)

so it is clear that we need a different approach and we are lucky becasue blacka and merton hyponthesied what we should do in this case we will create a replicating portfolio. the only thing that this replicating portfolio must do is make sure that it that no matter what the stae is the value of the replicating portfolio must equal the value of the option so you can think of the replicating portfolio as a basket of primary assets which has the value of the option in all states of world

\begin{align} \psi \longrightarrow number of shares \beta \longrightarrow number of bank \end{align}

(14)

so in order to determine these values and in order to abbide by the above condition we form the below equations

in the good state

\begin{align} 140\phi + 1.06\beta =30 \end{align}

(15)

and in for the bad state we construct

\begin{align} 90\phi + 1.06\beta = 0 \end{align}

(16)

now that we have two equations with two unknowns we can solve for this (the determenant is positive etc)

solve solve solve

thus we end up with $\phi = 0.6$ and $\beta=-50.94$

(will suspend disbelief about buying 0.6 shares/fractional shares)

\begin{align} V_0 &=S_0\times\phi + B_0\times\beta = &=100\times\phi + 1\times\beta = &=100\times 0.6 + 1\times -50.94 = 9,06 \end{align}

(17)

so we have a portfolio $V$

and we know now that

\begin{align} V_0 = 9,06 \end{align}

(18)

\begin{align} V_T = 30 \end{align}

(19)

in the good state and

\begin{align} V_T = 0 \end{align}

(20)

in the bad state

thus by design the replicating portfolio V has the same value as the call option/derivative at expiry hence C_0 should be equal to V_0 or else there would be arbitrage, etc

so we have answred question one we now know that C_0 should be 9.06 and even more cool we have actaully also answered question 2 because in order to hedge C we can jut trade in the V i.e. the underlying assets

but what about those probablities, they didnt even feature even if we changes the probablit of g changed to 0.99 it would not make a difference and the price would stil be 9.06 thus this pricing process does not care about the real world probabilities

but we can still use probabilities they just have to be somewhat “special”

imagine we have

\begin{align} \mathbb{Q}(g) = q \end{align}

(21)

and

\begin{align} \mathbb{Q}(b) = 1-q \end{align}

(22)

and now we just have to “choose/find” q so that the replicating porfolio price V_0 is still consistent and V_0 = 9.06

thus we construct

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times S_{T}] \end{align}

(23)

which we call the expectation under the probability measure $\mathbb{Q}$

if we do construct this such that it equals the sahre price at time t=0

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times S_{T}] = S0 \end{align}

(24)

then we can use this in the way we tried to in (13) and udo expected present value

so lets try that

Undefined control sequence: \plus at position 109: …06^-1)\times q \̲p̲l̲u̲s̲ ̲(90\times1.06)^…

\begin{align}
 S_0  &= \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times S_{T}]
      &= (140\times 1.06^-1)\times q \plus (90\times1.06)^{-1})\times (1-q)
      &= (140\times 1.06^-1)\times q \plus (90\times1.06)^{-1})\times (1-q)
\end{align}

thus for the probability of being in a good state, we get

\begin{align} q = 0.32 \end{align}

(26)

and for the probability of being in a bad state we get

\begin{align} 1-q =0.68 \end{align}

(27)

finally we can apply this to the call option /derivati

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times C_{T}] &= solve &= solve &= 9.06 &= C_0 \end{align}

(28)

and we see tht indeed

So in summary the main points are

real world probabilites dont work on shares or options based on them and

this new probability measure {math}\mathbb{Q} has sme interesting properties

Unexpected end of input in a macro argument, expected '}' at end of input: ….06
\end{align}

\begin{align}
\mathbb{E}^{\mathbb{Q}}(\frac{C_T]{C_0}) = \mathbb{E}^{\mathbb{Q}}(\frac{S_T]{S_0}) = \mathbb{E}^{\mathbb{Q}}(\frac{B_T]{B_0}) = 1.06
\end{align}

thus all things/ assets grow at the risk free rate and hence this probability measure Q is know as the Risk-Neutral Measure

it is also known as the the pricing measure or equivalent martinggale measure

but the P and Q are equivalent in that they:

assign non zero probabilities to the same sets even though those probabilitys may be different
assign zero probability to the same sets

Also any asset or “thing” that abides byt the discounted expectation under Q like

the share

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times S_{T}] &= S_0 \end{align}

(30)

the call option

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times C_{T}] &= C_0 \end{align}

(31)

any other thing

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times G_{T}] &= G_0 \end{align}

(32)

is called a martingale

and similarly if we have a thing J and we know J_T and we are told/ know its a martingale. we can do

\begin{align} \mathbb{E}^{\mathbb{Q}}[1.06^{-1} \times J_{T}] &= J_0 \end{align}

(33)

to find J_0

hence the equivalent martinggale measure naming hence you could say measure Q turns discounted asset prices into martingales

does this measure always exist the answer is no sometiems it doesnt

but then what do you do well you can’t do anything

the fundamental theorem of asset prcinging part 1 states and EMM means there is NO ARBITRGE (Harriosn And Krepps)

so when there is no EMM there is arbitrage but when there is more than one EMM you can easily show that there is infinitely many EMMS and this is termed an incomplete market and it means not all derivates can be replicated so some derivates can but not all

the fundamental theorem of asset prcinging part 2 states if NA is true then a market is complete complete and every derivative can be hedge if and only iff there is 1 EMM

Black Scholes¶

squigggely line

continuous stochastic process bachalier phd