Probability and Statistics

\begin{align}
E[T_0|S_0 = n) &= E[T_{n-1}|S_0=n] + E[T_0|S_0=n-1]\\
&= E[T_{n-1}|S_0=n] + (n-1)E[T_0|S_0=1]\\
&= E[T_{0}|S_0=1] + (n-1)E[T_0|S_0=1]\\
\end{align}

BAYMAX

how from first t osecond step?

like factoring n-1 out?

Siong Thye Goh

induction

write down the induction hypothesis and u have a tool to use

BAYMAX

it sems like we are using the result we want t o prove

Siong Thye Goh

we want to prove for case n, we assume it is true for n-1

BAYMAX

2:49 PM

Oh man yes

right right

Oh so we have it nice nice

$E(T_{0}|S_{0}=1)$

means the expected time

reuired for the wealth at 1 to reach 0

?

Why they would be different for two different cases like $p<q$ and $p \geq q$

Siong Thye Goh

i m guessing (i have not computed anything though i should know these results)

one of them is going to be infinity

and u can guess which one

=)

but i m uncertain as i have not done such things for years

BAYMAX

$p \geq q$

Siong Thye Goh

try to compute the value to find out?

BAYMAX

?

Siong Thye Goh

i m guessing but i m not sure

if one of them is infinity, it has to be that one

BAYMAX

2:55 PM

But if i use the result we obtained

$E(T_{0}|S_{0}=i) = i E(T_{0}|S_{0}=1)$

i just put $i=1$

to get the LHS =RHS

from state i to state 0

that is from state 1 t ostate zero

Siong Thye Goh

u have to think of how to use $p$ and $q$ somewhere

BAYMAX

$S_{1} = S_{0}+1$ with probability $p$

$S_{1} = S_{0} - 1$ with probability $q$

Siong Thye Goh

yes

u r moving in the right direction

BAYMAX

Hm i m not usre wht next

Siong Thye Goh

$$E[T_0|S_0=1]=q+pE[T_0|S_0=2]$$

BAYMAX

3:08 PM

Still trying :'(

what is the expression for p an d q?

Siong Thye Goh

take ur time

?

p is the probability to move to the right

q is the probability to move to the left right

i think i might have made a mistake

$$E[T_0|S_0=1] = q + p(1+E(T_0|S_0=2))$$

BAYMAX

$E(T_{0}|S_{0}=2) = E(T_{1}|S_{0}=2)+E(T_{0}|S_{0}=1)$

$E(T_{1}|S_{0}=2) = q$

as moving from state 2 to 1

Siong Thye Goh

actually the equation that i gave u... u just have to solve for $E[T_0|S_0=1]$

$E[T_1|S_0=2] = E[T_0|S_0 =1 ]$

BAYMAX

is it like this

we go from state 1 to 0

we can go thru 2 ways

1 t o0

Siong Thye Goh

u can directly go to $0$ or u can go to $2$ first

that is what the equation I wrote above means

BAYMAX

3:19 PM

yes

Siong Thye Goh

$$E[T_0|S_0=1] = q + p(1+E(T_0|S_0=2))$$

BAYMAX

1) going from state 1 to state 0

2) state 1 to stae 2 then stat 0

?

Siong Thye Goh

yup

BAYMAX

yes stae 1 to 0 is q

now from state 2 t ostate 0

?

why p is involved?

eventhough we are going left

that is 2 to 0

Siong Thye Goh

from state 1 to 0 take time 1 with probability q

BAYMAX

3:22 PM

yes

Siong Thye Goh

i mean with probability q, we go from state 1 to state 0, and that would take 1 unit

and with probability p, we go from state 1 to state 2 , and that would take ... the expression that i typed earlier

BAYMAX

state1 to stae2

is p

state 2 to state 0

?

Siong Thye Goh

the probability is p, and the time needed would be $1+E(T_0|S_0=2)$

we use the law of total expectation

*away a while

BAYMAX

wll be back soon

Siong Thye Goh

3:46 PM

back

BAYMAX

me too

oh ok

How about $p<q$, $p \geq q$

?

cases

Siong Thye Goh

the equation that i wrote

solve for $E[T_0|S_0=1]$

BAYMAX

like in question it has divided into cases of the above expectatio

Siong Thye Goh

solving that would give u a big hint

BAYMAX

q/1-2p

Siong Thye Goh

4:02 PM

hmmm

somehow i get $1/(1-2p)$

but yup, it's just algebra now

at least our denominator is the same

so now u can see what happens when $p<\frac12$ and what happens otherwise

BAYMAX

yes it must be $\frac{1}{1-2p}$

For $p=0.5$, it blows up

Siong Thye Goh

yup

BAYMAX

p<0.5, it simply is positive

p>0.5 its negative not possile

*possible

Siong Thye Goh

yup, it means it's also infinity

BAYMAX

oh ok

How we related p and q

q was independent here

Siong Thye Goh

4:10 PM

$p+q=1$ is n't it?

BAYMAX

yes

Siong Thye Goh

$p \ge q \iff p \ge \frac12$

BAYMAX

oh yes

Oh so our past first gor over

its deep no

Siong Thye Goh

erm... can't comprehend what was typed

BAYMAX

Like it has deep concepts involved

can u plz give solutions to lst two parts, i will try to understnad?

Siong Thye Goh

4:17 PM

the strategy is the same

induction

BAYMAX

its much time since i used stochastics

so i said

Here we are mulitplying

Siong Thye Goh

yes, it is by induction

induction is not relying on addition

BAYMAX

?

induction is not relying on addition?

$P(T_{0}<\infty|S_{0}=i)$

tells that probability to move from state $i$ to some finite state

Siong Thye Goh

use the idea that u ahve to go throught state i-1

if u understand the first part u should be able to do this as well

(whcih explians why only 1 mark)

BAYMAX

let me work on it

i will let u know

by ping

its night

hv t oeat hungry nd sleep

Siong Thye Goh

4:25 PM

eat more

BAYMAX

yeah

:)

Ok bye

i will discuss tmrw

Siong Thye Goh

bye

Transcript for

$Mathematics$ Probability and Statistics