Distribución de

Suponer que $X$ tiene la distribución beta Beta $(1,K-1)$ y $Y$ sigue un chi-cuadrado con $2K$ grados Además, suponemos que $X$ y $Y$ son independientes

¿Cuál es la distribución del producto? $Z=XY$ .

Actualizar
mi intento:

\begin{aligned} f_{Z} & = \int_{y = - \infty}^{y = + \infty} \frac{1}{| y |} f_{Y} (y) f_{X} (\frac{z}{y}) d y \\ = \int_{0}^{+ \infty} \frac{1}{B (1, K - 1) 2^{K} Γ (K)} \frac{1}{y} y^{K - 1} e^{- y / 2} (1 - z / y)^{K - 2} d y \\ = \frac{1}{B (1, K - 1) 2^{K} Γ (K)} \int_{0}^{+ \infty} e^{- y / 2} (y - z)^{K - 2} d y \\ = \frac{1}{B (1, K - 1) 2^{K} Γ (K)} [- 2^{K - 1} e^{- z / 2} Γ (K - 1, \frac{y - z}{2})]_{0}^{\infty} \\ = \frac{2^{K - 1}}{B (1, K - 1) 2^{K} Γ (K)} e^{- z / 2} Γ (K - 1, - z / 2) \end{aligned}

$\begin{align} f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{0}^{+\infty} \frac{1}{B(1,K-1)2^K \Gamma(K)} \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{0}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} [-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})]_0^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1,-z/2) \end{align}$

¿Es correcto? en caso afirmativo, ¿cómo llamamos a esta distribución?

probability self-study distributions random-variable beta-distribution tam
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Si se trata de tarea o de autoaprendizaje, agregue la etiqueta correspondiente. No (usualmente) resolvemos tales problemas por usted, sino que lo ayudamos a guiarlo hacia una solución usted mismo, lo que en general le dará una mejor comprensión de cómo resolver dichos problemas en el futuro.

jbowman

No estoy seguro, pero tal vez esto sea de alguna ayuda: en.wikipedia.org/wiki/Noncentral_beta_distribution

¿Has intentado crear una segunda variable? Decir

W = X + Y

$W=X+Y$ ? Entonces podría obtener la distribución conjunta de

W, Z

$W,Z$ e integrarse

W

$W$ para obtener la distribución de

Z

$Z$ .

No veo dónde está utilizando el hecho de que la función de densidad Beta es cero en el complemento del intervalo

[0, 1]

$[0,1]$ .

whuber

@whuber Creo que encontré el error. ¿Desea dar una respuesta completa o lo hago yo solo?

tam

Respuestas:

Después de algunos comentarios valiosos, pude encontrar la solución:

We have $f_X(x)=\frac{1}{B(1,K-1)} (1-x)^{K-2}$ and $f_Y(y)=\frac{1}{2^K \Gamma(K)} y^{K-1} e^{-y/2}$ .

Also, we have $0\le x\le 1$ . Thus, if $x=\frac{z}{y}$ , we get $0 \le \frac{z}{y} \le 1$ which implies that $z\le y \le \infty$ .

Hence:

\begin{aligned} f_{Z} & = \int_{y = - \infty}^{y = + \infty} \frac{1}{| y |} f_{Y} (y) f_{X} (\frac{z}{y}) d y \\ = \int_{z}^{+ \infty} \frac{1}{B (1, K - 1) 2^{K} Γ (K)} \frac{1}{y} y^{K - 1} e^{- y / 2} (1 - z / y)^{K - 2} d y \\ = \frac{1}{B (1, K - 1) 2^{K} Γ (K)} \int_{z}^{+ \infty} e^{- y / 2} (y - z)^{K - 2} d y \\ = \frac{1}{B (1, K - 1) 2^{K} Γ (K)} {[- 2^{K - 1} e^{- z / 2} Γ (K - 1, \frac{y - z}{2})]}_{z}^{\infty} \\ = \frac{2^{K - 1}}{B (1, K - 1) 2^{K} Γ (K)} e^{- z / 2} Γ (K - 1) \\ = \frac{1}{2} e^{- z / 2} \end{aligned}

$\begin{align} f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{z}^{+\infty} \frac{1}{B(1,K-1)2^K \Gamma(K)} \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{z}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} \left[-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})\right]_z^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1) \\ &= \frac{1}{2} e^{-z/2} \end{align}$ where the last equality holds since

B (1, K - 1) = \frac{Γ (1) Γ (K - 1)}{Γ (K)}

$B(1,K-1)=\frac{\Gamma(1)\Gamma(K-1)}{\Gamma(K)}$ .

So $Z$ follows an exponential distribution of parameter $\frac{1}{2}$ ; or equivalently, $Z \sim\chi_2^2$ .

tam
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There is a pleasant, natural statistical solution to this problem for integral values of $K$ , showing that the product has a $\chi^2(2)$ distribution. It relies only on well-known, easily established relationships among functions of standard normal variables.

When $K$ is integral, a Beta $(1,K-1)$ distribution arises as the ratio

\frac{X}{X + Z}

$\frac{X}{X+Z}$ where

X

$X$ and

Z

$Z$ are independent,

X

$X$ has a

χ^{2} (2)

$\chi^2(2)$ distribution, and

Z

$Z$ has a

χ^{2} (2 K - 2)

$\chi^2(2K-2)$ distribution. (See the Wikipedia article on the Beta distribution for instance.)

Any $\chi^2(n)$ distribution is that of the sum of squares of $n$ independent standard Normal variates. Consequently, $X+Z$ is distributed as the squared length of a $2 + 2K-2 = 2K$ vector with a standard multinormal distribution in $\mathbb{R}^{2K}$ and $X/(X+Z)$ is the squared length of the first two components when that vector is radially projected to the unit sphere $S^{2K-1}$ .

The projection of a standard multinormal $n$ -vector onto the unit sphere has a uniform distribution because the multinormal distribution is spherically symmetric. (That is, it is invariant under the orthogonal group, a result that follows immediately from two simple facts: (a), the orthogonal group fixes the origin and by definition does not change covariances; and (b) the mean and covariance completely determine the multivariate normal distribution. I illustrated this for the case $n=3$ en https://stats.stackexchange.com/a/7984 ). De hecho, la simetría esférica muestra inmediatamente que esta distribución es uniforme condicional a la longitud del vector original. El radio $X/(X+Z)$ por lo tanto es independiente de la longitud.

Lo que todo esto implica es que multiplicar $X/(X+Z)$ por un independiente $\chi^2(2K)$ variable $Y$ creates a variable with the same distribution as $X/(X+Z)$ multiplied by $X+Z$ ; to wit, the distribution of $X$ , which has a $\chi^2(2)$ distribution.

whuber
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Very nice analogy! I feel a bit uncertain about the final paragraph though as the simplification only occur because

X + Z

$X+Z$ is on both sides of the multiplication, which cannot work for an independent

χ^{2} (2 K)

$\chi^2(2K)$ .

Xi'an

But after some further musing in the Paris métro, I realised that because

X / (X + Z)

$X/(X+Z)$ and

(X + Z)

$(X+Z)$ are independent, using

(X + Z) \times X / (X + Z)

$(X+Z)\times X/(X+Z)$ or using

Y \times X / (X + Z)

$Y \times X/(X+Z)$ lead to the same distribution. Congrats!

Xi'an

addendum: the reasoning goes for non-integer K's as well, if one defines a

χ_{q}^{2}

$\chi^2_q$ as a Gamma

Ga (q / 2, 1 / 2)

$\text{Ga}(q/2,1/2)$ .

Xi'an

@Xi'an Thank you for those revealing comments. Indeed, one way to exploit the recognition that

X / (X + Z)

$X/(X+Z)$ and

X + Z

$X+Z$ are independent is to pursue the implication that their density functions will be separable: and that idea applies without modification to the general case of non-integral

K

$K$ . Even for those who prefer to compute the convolution

X Y

$XY$ directly, these statistical insights suggest a simple and effective way of proceeding with the integration by means of an appropriate change of variables.

whuber

I greatly deprecate the commonly used tactic of finding the density of $Z = g(X,Y)$ by computing first computing the joint density of $Z$ and $X$ (or $Y$ ) because it is "easy" to use Jacobians, and then getting $f_Z$ as a marginal density (cf. Rusty Statistician's answer). It is much easier to find the CDF of $Z$ directly and then differentiate to find the pdf. This is the approach used below.

$X$ and $Y$ are independent random variables with densities $f_X(x) = (K-1)(1-x)^{K-2}\mathbf 1_{(0,1)}(x)$ and $f_Y(y) = \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2}\mathbf 1_{(0,\infty)}(y)$ . Then, with $Z = XY$ , we have for $z > 0$ ,

\begin{aligned} P {Z > z} & = P {X Y > z} \\ = \int_{y = z}^{\infty} \frac{1}{2^{K} (K - 1)!} y^{K - 1} e^{- y / 2} [\int_{x = \frac{z}{y}}^{1} (K - 1) (1 - x)^{K - 2} d x] d y \\ = \int_{y = z}^{\infty} \frac{1}{2^{K} (K - 1)!} y^{K - 1} e^{- y / 2} {(1 - \frac{z}{y})}^{K - 1} d y \\ = \int_{y = z}^{\infty} \frac{1}{2^{K} (K - 1)!} (y - z)^{K - 1} e^{- y / 2} d y \\ = e^{- z / 2} \int_{0}^{\infty} \frac{1}{2^{K} (K - 1)!} t^{K - 1} e^{- t / 2} d y on setting y - z = t \\ = e^{- z / 2} on noting that the integral is thatof a Gamma pdf \end{aligned}

$\begin{align} P\{Z > z\} &= P\{XY > z\}\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2} \left[\int_{x=\frac{z}{y}}^1 (K-1)(1-x)^{K-2}\,\mathrm dx \right] \,\mathrm dy\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2} \left(1-\frac{z}{y}\right)^{K-1}\,\mathrm dy\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} (y-z)^{K-1} e^{-y/2} \,\mathrm dy\\ &= e^{-z/2}\int_0^\infty \frac{1}{2^K (K-1)!}t^{K-1} e^{-t/2} \,\mathrm dy~~~\scriptstyle{\text{on setting}~y-z = t}\\ &= e^{-z/2}\qquad\scriptstyle{\text{on noting that the integral is that of a Gamma pdf}}\\ \end{align}$

It is well-known that if $V \sim \mathsf{Exponential}(\lambda)$ , then $P\{V > v\} = e^{-\lambda v}$ . It follows that $Z = XY$ has an exponential density with parameter $\lambda = \frac 12$ , which is also the $\chi^2(2)$ distribution.

Dilip Sarwate
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