Question

if p=(1+r-d)/(u-d) ---eq(1)
and
q=(u-1-r)/(u-d) ---eq(2)
then how does it satisfies (pu+qd)/(1+r)=1 ?
how can we make (pu+qd)/(1+r)=1 from equations (1) & (2) ?????

Answer 1

also q=1-p

Answer 2

@KingGeorge @satellite73

Answer 3

@amistre64

Answer 4

is this from probability? where do these come from?

Answer 5

@satellite73 you can ignore that q=1-p,
just the question is:
if p=(1+r-d)/(u-d) ---eq(1) and q=(u-1-r)/(u-d) ---eq(2) then how does it satisfies (pu+qd)/(1+r)=1 ? how can we make (pu+qd)/(1+r)=1 from equations (1) & (2) ?????

Answer 6

\[\frac{1+r-d}{u-d}+\frac{u-1-r}{u-d}=\frac{u-d}{u-d}=1\] in any case

Answer 7

I don't understand that where has (pu+qd)/(1+r)=1 come from

Answer 8

it's in stochastic calculus

Answer 9

I think I've got it. First, note that \[pu-pd=1+r-d\]and\[qu-qd=u-r-1\]Solve for \(pu\) and \(qd\), and add them together to get \[pu+qd=1+r+1+r-d-u+pd+qu\]Substitute \(q=1-p\) in that last q there to get\[pu+qd=1+r+1+r-d-u+pd+(1-p)u\]Simplify, to get \[pu+qd=1+r+1+r-d+pd-pu\]Notice that \(pd-pu=-(1+r-d)\). Do the substitution, and you get\[pu+qd=1+r\]The relation you want immediately follows.

Answer 10

thanks @KingGeorge