Distribución de

8

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?

— tam
fuente

2

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

$Z$ .

1

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

9

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
fuente

8

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
fuente

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

1

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

1

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

1

@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

3

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
fuente