Modes of a Bimodal Distribution

Question

I recently came across a problem in which I had to show that a discrete random variable $X$ has two modes $m_1$ and $m_2$. The information given was that $$\frac{P(X=n)}{P(X=n-1)} = \frac{0.9(n-1)}{n-3}$$

I need to show that this distribution has two modes, $m_1$ and $m_2$. My initial thoughts were to try and find the probability distribution, but could not do so from the data above. I don't really know how to approach this problem so any inputs would be appreciated.

Answer 1

Observe that: $$\frac{P(X=21)}{P(X=20)}=\frac{0.9(21-1)}{21-3}=1$$ or equivalently: $$P(X=20)=P(X=21)$$

Now it remains to show that: $$n\notin\{20,21\}\implies P(X=n)<P(X=20)=P(X=21)$$

If $n\geq22$ then the RHS is less than $1$ so that $P(X=n-1)>P(X=n)$ implying that: $$P(X=21)>P(X=22)>P(X=23)>\cdots$$

If $n\leq20$ then the RHS exceeds $1$ so that $P(X=n)>P(X=n-1)$ implying that: $$P(X=20)>P(X=19)>P(X=18>\dots$$

Answer 2

We have the recurrence $$ \mathbb P(X=n) = \frac{9(n-1)}{10(n-3)}\mathbb P(X=n-1).\tag1 $$ Since probabilities cannot be negative, it follows that $\mathbb P(X\geqslant 3)=1$. Let $p_n=\mathbb P(X=n)$, then $$p_n = C\left(\frac59\right)^{3-n}2^{2-n}(n-2)(n-1) $$ for some constant $C$ (as can be verified by induction). Since $\sum_{n=3}^\infty p_n=1$, we have $$ C = \left(\sum_{n=3}^\infty \left(\frac59\right)^{3-n}2^{2-n}(n-2)(n-1)\right)^{-1} = \frac1{1000}, $$ and hence $$ p_n = \frac{(n-2)(n-1)}{1458}\left(\frac9{10}\right)^n. $$ Let $g(x) = \frac{(x-2)(x-1)}{1458}\left(\frac 9{10}\right)^x$ for $x\geqslant 3$, then $g'(x)=0$ has approximate solution $20.4956$. Since $g''(20.4956)\approx-0.000158506<0$, it follows that this is a local maximum of $g$. So the integer values of $n$ that maximize $g$ are $20$ and $21$, with $$ p_{20} = p_{21} = \frac{2851798070642983299}{100000000000000000000}\approx 0.028518. $$