How do I prove that the following discrete distribution is unimodal (in the sense that it has only one local maxima)? \begin{align} & P(X=k)=\left. \frac{1}{k!}\frac{{{d}^{(k)}}}{dz}\prod\limits_{i=1}^{n}{(p{_i}+q{_i}\cdot z+r{_i}\cdot {{z}^{2}})\ } \right|\ z=0 \ & for\quad k:0,1,...,2n \ \end{align} for any $n\ge 1$, where for each $i ,\quad 0 and it is never the case that $p{_i}>q{_i}
Unimodal discrete distribution
0
$\begingroup$
probability
-
2You take a product over $i$, but there's no $i$ there. Is that what you want? – 2012-02-15
-
0In general, look at the _ratio_ $\frac{P\{X=k\}}{P\{X=k-1\}}$ as a function of $k$. Typically, you will find that the _ratio_ $> 1$ for $k \leq M$ and the _ratio_ $ < 1$ for $k > M$; meaning that $P\{X=k\}$ as a function of $k$ increases as $k$ increases, had a peak at $k = M$ and decreases thereafter. So $M$ is the location of the mode. – 2012-02-15
-
0Is the question being asked the following: Let $Y_i$, $ 1 \leq i \leq n$, be $n$ i.i.d. discrete random variables taking on values $0, 1, 2$ with probabilities $p$, $q$, and $r=1-p-q$ respectively, and let $X = \sum_{i=1}^n Y_i$. Prove that the probability mass function of $X$ is unimodal. – 2012-02-15
-
0It would be more usual to have $0\le p,q,r$ and $p+q+r=1$ if these are to be probabilities. It does not really matter, since it only produces a constant, which does not affect unimodality. – 2012-02-15
-
0@Gerry: If Dilip's second comment is correct, then this is a trinomial RV so you can replace the product with a power. – 2012-02-15
-
0Thanks @Dilip - I corrected the question, please follow-up. – 2012-02-15
-
0@Dilip, I added another condition... – 2012-02-15
-
0$(p,q,r) = (0.7,0.2,0.1)$ is such that the condition $p > q < r$ does not hold. Is that acceptable, or do you want $q > \max\{p,r\}$ to hold so that $q$ is the mode? – 2012-02-15
-
0No, my condition is sufficient (q does not need to be the mode of the distribution at n=1). – 2012-02-15