# Identification of Probability Distribution

#### masood

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
It looks like a binomial under a condition.

#### masood

Thanks for the reply. How can one know if it is conditioned ?

#### masood

@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
departures are independent
can't be right. Otherwise, your final result would be correct. You need to work out the dependence.

#### masood

@mathman Can you help me on this, how get expression for $P(D=d)$ and $P(\bar{D}=\bar{d})$

#### mathman

Forum Staff
I don't fully understand the scenario.