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Suppose that $X$ is distributed as a Binomial$(n,p)$ and $Y$ as a Binomial Negatively$(r,p)$. Prove that $$F_X(r-1) = 1 - F_Y(n-r)$$ For $X \sim Bin(n,p)$,
$$F_X(x) = P(X \le x) = \sum_{i=0}^{x} \binom{n}{i} p^i (1-p)^{n-i}$$ And for $Y \sim BN(r,p)$, $$F_Y(y) = P(Y \le y) = \sum_{j=0}^{y} \binom{r+j-1}{j} p^r (1-p)^j$$ Starting from the right side of the equality: \begin{align*} 1 - F_Y(n-r) &= 1 - P(Y \le n-r) \\ &= 1 - (1 - P(Y > n-r)) \\ &= P(Y > n-r) \\ &= \sum_{j=n-r+1}^{\infty} \binom{r+j-1}{j} p^r (1-p)^j \end{align*} Now, let $j = n - r + 1 + k$, so $k$ starts at $k=0$. Thus, let $k = j - (n - r + 1)$, then this new index starts at $0$, and we have: \begin{align*} \sum_{j=n-r+1}^{\infty} \binom{r+j-1}{j} p^r (1-p)^j &= \sum_{k=0}^{\infty} \binom{n+k}{k} p^r (1-p)^{n-r+1+k} \\ &= p^r (1-p)^{n-r+1} \sum_{k=0}^{\infty} \binom{n+k}{k} (1-p)^{k} \\ &= p^r (1-p)^{n-r+1} \\ \end{align*}

What would be the next step, I'm stuck with this last identity. Maybe I should use convergence???

RobPratt
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Roma_Rayado
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2 Answers2

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The algebraic approach you take has been posted to math.SE before, e.g., see this question. However, I think this is a clunky way to look at this property, and there's a much nicer conceptual proof, which I'll go over below (this has probably also been posted before, but I've already written it out :)).

Recall that $\mathrm{Bin}(n,p)$ is the distribution of the number of successes you see if you toss $n$ coins independently, each of which has probability of success $p$. Further, $\mathrm{NegBin}(r,p)$ is the distribution of the number of failures you see before observing $r$ successes if you toss $p$-coins independently (see below if you haven't seen this definition previously).

Now, let $X_1, \cdots, X_t, \cdots$ be an infinite iid stream of $p$-coin flips, and let $S_t$ be the running sum $S_t = \sum_{1\le s \le t}X_s$. Let $\tau := \inf\{t : S_t = r\}.$ Notice then that $\tau - r$ is distributed as a negative binomial: since the $r$th success is at $\tau$, so the total number of failures is $\tau - r$. Finally, observe that $S_t$ is a nondecreasing process (so $S_t \le r \implies S_{t-k} \le r$ for any $k \ge 0$). Therefore, $$ P(\tau > t) = P(\forall s \le t, S_s < r) = P(S_t \le r-1) = F_{\mathrm{Bin}(t,p)}(r-1) $$

But, since $\tau - r$ is a negative-binomial, $$1 - F_{\mathrm{NegBin}(r,p)}(t-r) = P(\tau - r > t-r) = P(\tau > t) = F_{\mathrm{Bin}(t,p)}(r-1),$$ and this holds for all naturals $t,r,$ and any $p \in [0,1]$.

In principle, depending on whether you have seen this or not, to complete the conceptual proof, we'd need to show that $F_{\mathrm{NegBin}(r,p)}(t-r) = P(\tau \le t) \iff P_Y(t-r) = P(\tau = t).$ To see this, notice that $\tau = t$ means that the $t$th toss was a heads, and there were $r-1$ heads in the first $t-1$ tosses. This occurs with chance $$ p \cdot \binom{t-1}{r-1} p^{r-1} (1-p)^{t-r} = \binom{t -1}{t-r}p^r(1-p)^{t-r} = \binom{k +r-1}{k}p^r(1-p)^k,$$ where $k = t-r.$

stochasticboy321
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The proof with words is:

  • $F_X(r-1)$ is the probability you see $r-1$ or fewer successes in $n$ attempts,

  • which is equivalent to the probability of seeing strictly fewer than $r$ successes and strictly more than $n-r$ failures (in $n=r+(n-r)$ attempts),

  • while $1 - F_Y(n-r)$ is the complement of the probability you see $n-r$ failures or fewer by the time you see the $r$th success,

  • i.e. it is the probability you see strictly more than $n-r$ failures by the time you see the $r$th success,

  • so these are the same probabilities.

Henry
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