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negative binomial random variable works like this, but can you explain why we use ${n-1 \choose r-1}$ instead of ${n \choose r}$?

$P(X=x) = {x-1\choose r-1}(1-p)^{x-r}p^r$

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Basic reason the use of ${n-1} \choose {r-1} $ lies in the description of the situation where negative binomial occurs.We want rth success in nth trial.Hence,$r-1$ success in $n-1$ trial and nth trial being a success.Hence $n-1 \choose r-1 $.

Now you can redefine the situation in which $n \choose r$ makes sense but try doing it,it is not very easy to perceive.

EDIT:As asked in comments,little elaboration.

Lets start from scratch: We want first $n-1$ trials result in $r-1$ successes and $(n-1)-(r-1)=n-r$ failures.Any of $r-1$ out of $n-1$ trials can be success,and remaining $n-r$ should be be failures (note that choosing $r-1$ success also fixes $n-r$ failures.Hence ${n-1 \choose r-1} \times p^{r-1} \times {(1-p)}^{n-r}$ . Now the last trial must be success,probability of which is $p$,hence we multiply by $p$ and get required answer.

Hope this helps.

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    No we are not skipping the last trial.However note that last trial is **not** guaranteed success,it is success with probability $p$.Hence we multiplied by $p$,otherwise we'ld have got ${n-1 \choose r-1} {(1-p)}^{(n-r)}p^{(r-1)}$2012-10-04
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Dunno, maybe for the sum of independent NB$(r,p)$ and NB$(s,p)$ to be NB$(r+s,p)$.

Slightly less convincing: for NB$(r,p)$ to be geometric when $r=1$.