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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 0q{_i}

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    You take a product over $i$, but there's no $i$ there. Is that what you want?2012-02-15
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    In 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
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    Is 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
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    It 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
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    @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
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    Thanks @Dilip - I corrected the question, please follow-up.2012-02-15
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    @Dilip, I added another condition...2012-02-15
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    $(p,q,r) = (0.7,0.2,0.1)$ 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
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    No, my condition is sufficient (q does not need to be the mode of the distribution at n=1).2012-02-15

2 Answers 2

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If there are no further conditions on $p$, $q$ and $r$ apart from $0

Let $n=2$, so $\prod\limits_{i=1}^{n}{(p+q\cdot z+r\cdot {{z}^{2}})\ } = p^2 + 2pq z +(q^2+2pr)z^2+ 2qr z^3+r^2z^4$ and the probabilities are $p^2, 2pq, q^2+2pr, 2qr, r^2$.

Now take $p=0.4$, $q=0.1$ and $r=0.5$. So the probabilities become $0.16,0.08,0.41,0.10,0.25$ and this is not a unimodal sequence.

Perhaps you need an additional condition that $p,q,r$ is unimodal.

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    I corrected the question, please follow-up.2012-02-15
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    Henry's counterexample still applies. (An even simpler version is $n=1$, same $p$, $q$, $r$.)2012-02-15
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    Thanks @Didier, I added the additional condition Henry implied.2012-02-15
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    @DidierPiau "even simpler version..." Actually the OP did say $n > 1$. But, I suppose unimodal could be taken to mean having a unique global maximum, which Henry's sequence does have, rather than monotone increasing up to the maximum and monotone decreasing thereafter.2012-02-15
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Work in progress Uncompleted proof of a special case of the desired result.

Let us call a sequence $(a_0, \ldots, a_n)$ as unimodal with mode $M$ if it enjoys the property that $$a_0 < a_1 < \cdots < a_{M-1} < a_M > a_{M+1} > \cdots > a_n$$ where $1 \leq M < n$.

Let $\{Y_i\}$, denote a sequence of be n i.i.d. discrete random variables taking on values $0,1,2$ with probabilities $p$, $q$, and $r=1−p−q$ respectively where we assume that $p < q > r$ so that $(p,q,r)$ is a unimodal sequence with mode $1$. Define $X_n = \sum_{i=1}^n Y_i$. Then, $X_n$ is a discrete random variable taking on value $i$ with probability $p_n(i)$, $0 \leq i \leq 2n$.

Claim: The sequence $(p_n(0), p_n(1), \ldots, p_n(2n))$ is unimodal.

We prove the claim by induction.

Basis: The result holds for $n = 1$ since we have assumed that $(p_1(0), p_1(1), p_1(2)) = (p,q,r)$ is unimodal.

Induction hypothesis: $(p_n(0), p_n(1), \ldots, p_n(2n))$ satisfies $$p_n(k) > p_n(k-1) ~\text{for}~ 1 \leq k \leq M_n, ~~ p_n(k-1) > p_n(k)~\text{for}~ M_n < k < 2n.$$

Now consider the sequence $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$. Since $$p_{n+1}(k) = p_n(k)\cdot p + p_n(k-1)\cdot q + p_n(k-2)\cdot r$$ (where we use $p_n(-2) = p_n(-1) = p_n(2n+1) = p_n(2n+2) = 0$), we have that for $1 \leq k \leq M_n$, $p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three positive terms and therefore $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(M_n))$ is an increasing sequence. Similarly, for $M_n +2 \leq k \leq 2n+2$, $p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three negative terms and therefore $(p_{n+1}(M_n+2), p_{n+1}(M_n+3), \ldots, p_{n+1}(2n+2))$ is a decreasing sequence.

  • If $p_{n+1}(M_n) < p_{n+1}(M_n+1) < p_{n+1}(M_n+2)$, then $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with mode $M_n+2$.

  • If $p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with mode $M_n+1$.

  • If $p_{n+1}(M_n) > p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with mode $M_n$.

All that remains to prove unimodality of $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is to show that the case $$p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$$ cannot occur. Compare to the OP's $p > q < r$ is not allowed.

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    Thanks, @Dilip. This is indeed (uncompleted) proof of a special case of my problem - what about the general case where only p>q2012-02-16
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    BTW, @Dilip, Your final double inequality line for the 4th possibility is a typo I think - it should read ...>...<..., in the form of p>q2012-02-16