Proving the Mode of Negative Binomial

Question

I refer to the answer by BruceET mentioned here: Finding the mode of the negative binomial distribution

It is mentioned that the mode of negative binomial distribution can be found by: t = 1 + ((r-1)/p) where t is some number, r is the number of successes and p is the probability of success If t is an integer, there will be 2 modes at t and t-1 If t is not an integer, the integer part of t is the mode. E.g. '3' is the integer part of 3.5

This is what I have worked on so far: ((r-1)/p) is the expected number of attempts to achieve all required successes excluding the final success The expected number of attempts + the definite final success = the total number of expected attempts.The total number of expected attempts is the mode.

Why could there be two modes of t-1 and t?

P(X=x) = ((x-1) choose (r-1))(p^r)(q^(x-r))

((t-2)choose(r-1))(p^r)(q^(t-r-1)) = ((t-1)choose(r-1))(p^r)(q^(t-r))

((t-2)choose(r-1))(q^-1) = ((t-1)choose(r-1))

((t-2)(t-3)...(t-r))/(r-1)! x (q^-1) = ((t-1)(t-2)...(t-r+1))/(r-1)!

(t-r)/q = t-1 [I am stuck here]

Furthermore, I don't know why if t is not an integer, the integer part of t is the mode.

Thanks very much

Answer 1

If you have $\dfrac{t-r}{q}=t-1$ then you also have $t-r=qt-q$ and thus $t=\dfrac{r-q}{1-q}= 1+\dfrac{r-1}{p}$.

This says that

• if $x < 1+\dfrac{r-1}{p}$ then $\mathbb{P}(X=x) < \mathbb{P}(X=x-1)$
• if $x = 1+\dfrac{r-1}{p}$ then $\mathbb{P}(X=x) = \mathbb{P}(X=x-1)$
• if $x > 1+\dfrac{r-1}{p}$ then $\mathbb{P}(X=x) < \mathbb{P}(X=x-1)$

So with $1+\dfrac{r-1}{p}$ integer then it and $\dfrac{r-1}{p}$ have the same highest probabilities and are the modes

while if $1+\dfrac{r-1}{p}$ is not an integer then the mode is the integer immediately below it