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Suppose $X_1,\ldots,X_n$ are each unimodal random variables with mode around zero. In particular, for each $i$, $X_i=X_i'-X'$ where $X'_i$ and $X'$ are i.i.d. unimodal random ($X_i'$ and $X_j'$ are ind. as well) variables so that $X_i$ is indeed unimodal. Is it the case that joint distribution of $X_i$ has the form, say $F(\cdot)$ satisfies the following:

Given $x_2,\ldots,x_n$ $F(u,x_2,\ldots,x_n)$ is convex for $u\in (-\infty,a)$ and concave for $u \in (a,\infty)$. In particular, the conditional (on truncating the other variables) distribution is unimodal. If so, how would one go about proving such a claim?

Thanks in advance

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3 Answers 3

I doubt it is unimodal about $0$. I would expect the conditional distribution to be unimodal about a value closer to the centre of the other values.

If $n$ is large and the $X'_i$ and $X'$ have unimodal symmetric distributions centred at $0$, I would expect the conditional distribution of $X_1$ given the values of $X_2,X_3,\ldots,X_n$ to be centred (and unimodal about) a value close to the centre of the values of $X_2,X_3,\ldots,X_n$ (which will be approximately the value of $-X'$).

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this was a reply to an earlier version of the question –  Henry Jul 6 '12 at 20:40

Notation: we have iid variables $Y,Y_1,\dots,Y_n$ and we observe $X_i=Y_i-Y$.

In general this is not true. If a distribution has a sharp peak at 0 and a heavy tail, then the a posteriori distribution on $Y$ caused by the observation of $x_1,\dots,x_n$ will have discrete peaks near $0,-x_1,\dots,-x_n$, rather than a single large peak at e.g. their empirical average (if $Y$ has a Gaussian distribution) or e.g. their median (if $Y$ has a Laplace distribution). So the distribution of $Y$ won't be unimodal, in the sense that its CDF won't be quasiconcave.

One might wonder if the sum $X_1=Y_1-Y$ might "blur" these peaks, but precisely because $Y_1$ has a sharp peak at 0 this is not going to change anything.

A simple example is, for $\delta<\tfrac{1}{2n}$ and for small $\varepsilon>0$, to let $Y$ be distributed: uniformly in $[-\delta/2,\delta/2]$ with probability $1-\varepsilon$, and uniformly in $[-1/2,1/2]$ with probability $\varepsilon$.

The corresponding density is $$\psi=(1-\varepsilon)\chi_{[-\delta/2,\delta/2]}+\varepsilon\chi_{[-1/2,1/2]}$$

Defining $x_0=0$, we have the equalities given by leonbloy in his answer: $$\psi(y) f(x_1x_2\dots x_n\mid y)=\prod_{i=0}^n \psi(x_i+y)\\ F(x_1x_2\dots x_n)=\int_{\mathbb R} \psi(y) f(x_1x_2\dots x_n\mid y)\,dy$$

So we have that when all $x_i$ (including $x_0$) are a distance greater than $\delta$ apart of each other: $$\psi(y) f(x_1x_2\dots x_n\mid y)=O(\varepsilon^{n-1})$$ $$F(x_1x_2\dots x_n) = O(\varepsilon^{n-1})$$ But when we relax the condition to allow $x_1=x_k$ (for some $k=0,2,3,\dots,n$), we have $$\psi(y) f(x_1x_2\dots x_n\mid y)=\frac{\varepsilon^{n-2}}{\delta^2} \Big[|x_k+y|\le\delta/2\Big]^2 + O(\varepsilon^{n-1})$$ $$F(x_1x_2\dots x_n) = \varepsilon^{n-2}/\delta + O(\varepsilon^{n-1})$$

So, for small enough $\varepsilon$ the conditional density of $X_1$ has $n$ peaks (near $x_0,x_2,\dots,x_n$) with up to $n-1$ valleys (this bound is reached when the peaks are more than $2\delta$ apart), therefore it cannot be unimodal when $n\ge 2$.

This counterexample can also be made smooth and made to have a unique mode by convolving it with a normal distribution of small variance, so the distribution is as well-behaved as we could hope for: it is symmetric, smooth and unimodal.

However, a weaker form of your question is true: if $Y$ has a log-concave distribution, then the a posteriori distribution of $Y$ will be log-concave (since the product of log-concave functions is log-concave), and so will $X_1$ (integrating a multivariate log-concave function along an axis yields a log-concave function, e.g. see theorem 1.8.3 of this book). So if I'm not mistaken your statement is true when replacing "unimodal" with "log-concave". This explains why the Gaussian and Laplace examples made sense!


Edit: Henry was wondering about the large $n$ limit case, for a fixed distribution. It is an interesting question that is true for a larger class of distributions, but I don't think it is true in general. For example if $Y$ is absolutely continuous with a density $\psi$ that diverges in 0 so that $\psi(y)^2$ is not integrable, then the density will diverge in each of $x_2,\dots,x_n$, so the conditional distribution is never unimodal unless $x_2=\dots=x_n$. A formal statement of the large $n$ limit problem could be: "Does the probability that $x_2,\dots,x_n$ satisfy the question converge to 1 as $n\to\infty$?" (according to the measure on $X_2\times\dots\times X_n$)

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For $x_2=x_3=\cdots=x_n=0$, do you think the statement is true? –  user25720 Jul 6 '12 at 20:53

Just some thougths. Let me change a little notation: we have $n+1$ iid variables $Y, Y_1, Y_2 \cdots Y_n$ with density $f_Y(y)=\psi(\cdot)$, and we are interested in the joint density of $X_i = Y_i - Y$.

We have $f_{X_1|Y}(x_1|y)=\psi(x_1+y)$ and they are jointly conditional independent. So

$$f(x_1 x_2 \cdots x_n |y)=\psi(x_1+y) \, \psi(x_2+y) \cdots \, \psi(x_n+y)=\frac{f(x_1 x_2 \cdots x_n )}{f_Y(y)}$$

Hence

$$ f_{X_1 X_2 \cdots X_n}(x_1 x_2 \cdots x_n ) = \int_{-\infty}^{+\infty} \psi(y) \,\psi(x_1+y) \, \psi(x_2+y) \, \cdots \, \psi(x_n+y) \, dy =\\ =\int_{-\infty}^{+\infty} \prod_{i=0}^n \psi(x_i+y) \, dy$$

where we've defined $x_0=0$. I'm not sure if much more can be said in general.

If $\psi(\cdot)$ is symmetric, then obviously this has the mode at $0$.

It's also true in general (but slightly trivial) that $E(X_1,X_2 \cdots X_n) = (0,0, \cdots 0)$. Further, it's also straighforward that $$E(X_1|X_2 \cdots X_n) = E(E( X_1 | X_2 \cdots X_n Y)) = \mu_y - E( Y | X_2 \cdots X_n) = \frac{\sum_{i=2}^n X_i}{n-1}$$

which agrees with Henry's observation.

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