Identification of Probability Distribution

Jan 2020
17
0
Pakistan
Hi
can any one tell what probaility distribution this equation refers to

'$p_i( \bar{D}=\bar{d} | D=d ) = {i \choose \bar{d}} {N_t -i \choose d-\bar{d}} \pi_1^d (1-\pi_1)^{N_t-d}$
 

mathman

Forum Staff
May 2007
6,933
775
It looks like a binomial under a condition.
 
Jan 2020
17
0
Pakistan
Thanks for the reply. How can one know if it is conditioned ?
 
Jan 2020
17
0
Pakistan
@mathman you are right it is conditioned and l need to understand derivation of this equation, it is for particular network slot (tagged slot) which is in state \(\displaystyle i\) (has currently \(\displaystyle i\) users ) . The probability of having \(\displaystyle \bar{D}=\bar{d}\) departures from it given that \(\displaystyle D=d\) departures occur in all available slots including tagged slot is given as
\(\displaystyle p_i( \bar{D}=\bar{d} | D=d ) = {i \choose \bar{d}} {N_t -i \choose d-\bar{d}} \pi_1^d (1-\pi_1)^{N_t-d}\) (given equation for derivation with \(\displaystyle 0 \leq \bar{d} \leq\) i ). \(\displaystyle Nt\) is total number of users that can be using this slot.

It is also told that the number of departures from all slots including tagged slot is \(\displaystyle d\). The departure of each users are independent events.

What i did for getting above result is shown below but i am not getting close to it.
since
\(\displaystyle Pi( \bar{D}=\bar{d} | {D}={d} ) = \frac{ P (\bar{D}=\bar{d}, {D}={d}) } { P({D}={d}) }\)
since departures are independent so I write it as

\(\displaystyle Pi( \bar{D}=\bar{d} | {D}={d} ) = \frac{ P (\bar{D}=\bar{d}) P( {D}={d}) } { P({D}={d}) }\)

it becomes

\(\displaystyle Pi( \bar{D}=\bar{d} | {D}={d} ) = P (\bar{D}=\bar{d}) \).

so i get following result by putting values in Bernoulli formula

\(\displaystyle Pi( \bar{D}=\bar{d} | {D}={d} ) = \sum_{\bar{d}=0}^i {i \choose \bar{d}} p^{ \bar{d} } (i-p)^{ i-\bar{d} }\).

I thought that Bernoulli formula may be used here

\(\displaystyle Pr[k\mbox{ successes in }n\mbox{ trials }] = \binom{n}{k}s^kf^{n-k}\).

or as

\(\displaystyle Pr[k\mbox{ successes in }n\mbox{ trials }] =\sum_{k=0}^n \binom{n}{k}s^kf^{n-k}\) .

but how can convert \(\displaystyle \sum\) into second \(\displaystyle { n \choose k }\) to derive given equation.


Am i right in making this decision ?. i am unable to derive it. Can any one tell me what i have done wrong?
 

mathman

Forum Staff
May 2007
6,933
775
departures are independent
can't be right. Otherwise, your final result would be correct. You need to work out the dependence.
 
Jan 2020
17
0
Pakistan
@mathman Can you help me on this, how get expression for $P(D=d)$ and $P(\bar{D}=\bar{d})$
 

mathman

Forum Staff
May 2007
6,933
775
I don't fully understand the scenario.