$Hello guys, how are you doing?

I've been doing some finance math and stuck at one moment.

Let's we have a price of a stock S(0) at the time 0. We are dividing the continuous timeline between 0 and N into discrete steps 0, 1, ... N.

At each step the price S(n) can go up and down with corresponding logarithmic returns ln(1+u) and ln(1+d). Thus at each step we have a variable k(n) which is distributed as Bernoulli distribution with two outcomes:

\begin{equation}
k(n) = \left\{
\begin{array}{ll}
ln(1+u) & \quad p\\
ln(1+d) & \quad 1-p
\end{array}
\right.
\end{equation}

where p=1-p

So if we consider all steps, we get a binomial tree, where each brunch represents possible movement of the stock's price S(0) up to S(N).

Now let's consider a random variable P = k(0) + ... + k(N). Each k(n) is i.i.d.

Its \mu is \sum_{n=1}^{N} E(k(n)) = N \cdot E(k(n)).
Its \sigma^2 is \sum_{n=1}^{N} var(k(n)) = N \cdot var(k(n))

Let's say that \tau = \frac{1}{N}

Hence, mean of each Bernoulli trial \mu{k(n)} is \mu \tau and \sigma of each trial is \sigma \sqrt{\tau}.

And here is a weird part - author then concludes that:

ln(1+u) = \mu\tau + \sigma\sqrt{\tau}
ln(1+u) = \mu\tau - \sigma\sqrt{\tau}

And it is not true i guess. According to his logic

ln(1+u) = mean + standard dev = p ln(1+u) + p ln(1+d) + p (ln(1+u)- \mu\tau)^2 + p(ln(1+d)-\mu\tau)^2

and the right part of the expression cannot be boiled down down to ln(1+u),because:
p ln(1+u) + p ln(1+d) + p (ln(1+u)- \mu\tau)^2 + p(ln(1+d)-\mu\tau)^2 =\= ln(1+u)

$

$Hello guys, how are you doing?

I've been doing some finance math and stuck at one moment.

Let's we have a price of a stock S(0) at the time 0. We are dividing the continuous timeline between 0 and N into discrete steps 0, 1, ... N.

At each step the price S(n) can go up and down with corresponding logarithmic returns ln(1+u) and ln(1+d). Thus at each step we have a variable k(n) which is distributed as Bernoulli distribution with two outcomes:

k(n) = \left\{
\begin{array}{ll}
ln(1+u) & \quad p\\
ln(1+d) & \quad 1-p
\end{array}
\right.

where p=1-p

So if we consider all steps, we get a binomial tree, where each brunch represents possible movement of the stock's price S(0) up to S(N).

Now let's consider a random variable P = k(0) + ... + k(N). Each k(n) is i.i.d.

Its \mu is \sum_{n=1}^{N} E(k(n)) = N \cdot E(k(n)).
Its \sigma^2 is \sum_{n=1}^{N} var(k(n)) = N \cdot var(k(n))

Let's say that \tau = \frac{1}{N}

Hence, mean of each Bernoulli trial \mu{k(n)} is \mu \tau and \sigma of each trial is \sigma \sqrt{\tau}.

And here is a weird part - author then concludes that:

ln(1+u) = \mu\tau + \sigma\sqrt{\tau}
ln(1+u) = \mu\tau - \sigma\sqrt{\tau}

And it is not true i guess. According to his logic

ln(1+u) = mean + standard dev = p ln(1+u) + p ln(1+d) + p (ln(1+u)- \mu\tau)^2 + p(ln(1+d)-\mu\tau)^2

and the right part of the expression cannot be boiled down down to ln(1+u),because:
p ln(1+u) + p ln(1+d) + p (ln(1+u)- \mu\tau)^2 + p(ln(1+d)-\mu\tau)^2 =\= ln(1+u)

$