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Let X, Y, Z be i.i.d. distributed Gamma random variables. What could the mode of the vector $(X, X+Y, X+Y+Z)$ be?

Does the mode of a random vector equal the combination of the marginal modes?

kjetil b halvorsen
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Jingjings
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  • The last question is interesting. You can construct counterexamples using bivariate Bernoulli variables (that is, $2\times 2$ contingency tables): try it! – whuber Nov 03 '13 at 23:34
  • The general answer to the second question is no\*. It's worth trying various simulations and some algebra. $\quad$ *I suspect (but haven't checked) that its not the case for your specific problem. – Glen_b Nov 03 '13 at 23:53
  • For multivariate mode see https://stats.stackexchange.com/questions/210107/whats-the-definition-of-multivariate-mode and https://stats.stackexchange.com/questions/33625/computationally-efficient-estimation-of-multivariate-mode – kjetil b halvorsen Apr 08 '21 at 04:40
  • https://stats.stackexchange.com/questions/91937 is also relevant (it concerns comparing marginal modes to the joint mode). – whuber Apr 08 '21 at 13:26

1 Answers1

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Let's address the questions in reverse order.

First, there isn't any general relationship between the marginal modes and the joint mode for arbitrary multivariate random variables. Counterexamples and explanations are given in the answers at How are the numbers of modes of marginal and joint distributions related?

Despite that, in the special circumstances of the first question (iid continuous variables), there is a nice, simple result. To obtain it, let $f$ be the common density function. The independence assumption means the joint density is their product,

$$f_{X,Y,Z}(x,y,z) = f(x)\,f(y)\,f(z).$$

In the change of variables $(X,U,V) = (X,X+Y,X+Y+Z),$ the probability element transforms as

$$\mathrm{d}x\,\mathrm{d}u\,\mathrm{d}v = \mathrm{d}x\,\mathrm{d}(x+y)\,\mathrm{d}(x+y+z) = \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z;$$

that is, it is unchanged. (The statistical argot is "the Jacobian equals $1$ everywhere." The actual calculation is performed with the wedge product as explained at https://stats.stackexchange.com/a/154298/919.) This is the key insight, because it means the maximum value of the joint distribution is obtained by maximizing

$$f_{X,U,V}(x,u,v) = f(x)\,f(u-x)\,f(v-u).$$

All such maxima obviously occur at points $(x,u,v)$ where $x,$ $u-x,$ and $v-u$ are modes of $f.$ Thus

The modes of the joint distribution of $(X,X+Y,X+Y+Z)$ are the points $(x^{*},x^{*}+y^{*},x^{*}+y^{*}+z^{*})$ where $x^{*},$ $y^{*},$ and $z^{*}$ are univariate modes of the common marginal distribution of $X,$ $Y,$ and $Z.$

When $f$ has a unique mode $x^{*},$ this implies $f_{X,U,V}$ has a unique mode at $(x^{*}, 2x^{*}, 3x^{*}).$ This is the situation of the question, because a $\Gamma(k,\theta)$ distribution ($\theta$ is a scale parameter) has a unique mode at $x^{*}=(k-1)\theta$ when $k\ge 1$ (and otherwise its mode is $x^{*}=0$).

It is apparent that results of this type are available for any number of iid continuous variables and any invertible linear transformation (because again the Jacobian is constant and so does not affect the joint mode).

whuber
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