Distribution du rapport entre deux variables aléatoires uniformes indépendantes

17

Supposons que $X$ et $Y$ sont standard uniformément distribués dans $[0, 1]$ , et ils sont indépendants, quel est le PDF de $Z = Y / X$ ?

La réponse d'un manuel de théorie des probabilités est

f_{Z} (z) = {\begin{cases} 1 / 2, & if 0 \leq z \leq 1 \\ 1 / (2 z^{2}), & if z > 1 \\ 0, & otherwise . \end{cases}

$f_Z(z) = \begin{cases} 1/2, & \text{if } 0 \le z \le 1 \\ 1/(2z^2), & \text{if } z > 1 \\ 0, & \text{otherwise}. \end{cases}$

Je me demande, par symétrie, ne doit pas $f_Z(1/2) = f_Z(2)$ ? Ce n'est pas le cas selon le PDF ci-dessus.

probability derived-distributions

— qed
source

Quel est le domaine de

et

?

X

$X$

Y

$Y$

— Sobi

en.wikipedia.org/wiki/Ratio_distribution

— kjetil b halvorsen

2

Pourquoi vous attendriez-vous à ce que cela soit vrai? La fonction de densité vous indique comment bien emballé la probabilité est dans le voisinage d'un point, et il est nettement plus difficile pour

soit près de

à

(pensez par exemple que

peut toujours être

, peu importe ce que

est, mais

quand

).

Z

$Z$

2

$2$

1 / 2

$1/2$

Z

$Z$

1 / 2

$1/2$

X

$X$

Z < 2

$Z < 2$

X > 1 / 2

$X > 1/2$

— dsaxton

2

Copie possible de Distribution d'un rapport d'uniformes: Qu'est-ce qui ne va pas?

— Xi'an

3

Je ne pense pas que ce soit un doublon, cette question est à la recherche du PDF, ici j'ai le PDF, je remets simplement en question son exactitude (peut-être plutôt naïvement).

— qed

19

La bonne logique est qu'avec indépendant , $X, Y \sim U(0,1)$ et $Z=\frac YX$ ont la mêmedistributionet donc pour $Z^{-1} =\frac XY$ $0 < z < 1$ où l'équation avec les CDF utilise le fait que

\begin{aligned} P {\frac{Y}{X} \leq z} & = P {\frac{X}{Y} \leq z} \\ = P {\frac{Y}{X} \geq \frac{1}{z}} \\ F_{Z} (z) & = 1 - F_{Z} (\frac{1}{z}) \end{aligned}

$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$

est une variable aléatoire continue et donc

. D'où le pdf de

satisfait

\frac{Y}{X}

$\frac YX$

P {Z \geq a} = P {Z > a} = 1 - F_{Z} (a)

$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$

Z

$Z$

f_{Z} (z) = z^{- 2} f_{Z} (z^{- 1}), 0 < z < 1.

$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$ Thus

f_{Z} (\frac{1}{2}) = 4 f_{Z} (2)

$f_Z(\frac 12) = 4f_Z(2)$ , and not

f_{Z} (\frac{1}{2}) = f_{Z} (2)

$f_Z(\frac 12) = f_Z(2)$ as you thought it should be.

— Dilip Sarwate
source

14

This distribution is symmetric--if you look at it the right way.

The symmetry you have (correctly) observed is that $Y/X$ and $X/Y = 1/(Y/X)$ must be identically distributed. When working with ratios and powers, you are really working within the multiplicative group of the positive real numbers. The analog of the location invariant measure $d\lambda=dx$ on the additive real numbers $\mathbb{R}$ is the scale invariant measure $d\mu = dx/x$ on the multiplicative group $\mathbb{R}^{*}$ of positive real numbers. It has these desirable properties:

$d\mu$ is invariant under the transformation $x\to ax$ for any positive constant $a$ :
$d μ (a x) = \frac{d (a x)}{a x} = \frac{d x}{x} = d μ .$ $d\mu(ax) = \frac{d(ax)}{ax} = \frac{dx}{x} = d\mu.$
$d\mu$ $x\to x^b$ $b$
$d μ (x^{b}) = \frac{d (x^{b})}{x^{b}} = \frac{b x^{b - 1} d x}{x^{b}} = b \frac{d x}{x} = b d μ .$ $d\mu(x^b) = \frac{d(x^b)}{x^b} = \frac{b x^{b-1} dx}{x^b} = b\frac{dx}{x} = b\, d\mu.$
$d\mu$ is transformed into $d\lambda$ via the exponential:
$d μ (e^{x}) = \frac{d e^{x}}{e^{x}} = \frac{e^{x} d x}{e^{x}} = d x = d λ .$ $d\mu(e^x) = \frac{de^x}{e^x} = \frac{e^x dx}{e^x} = dx = d\lambda.$ Likewise, $d\lambda$ is transformed back to $d\mu$ via the logarithm.

(3) establishes an isomorphism between the measured groups $(\mathbb{R}, +, d\lambda)$ and $(\mathbb{R}^{*}, *, d\mu)$ . The reflection $x \to -x$ on the additive space corresponds to the inversion $x \to 1/x$ on the multiplicative space, because $e^{-x} = 1/e^x$ .

Let's apply these observations by writing the probability element of $Z=Y/X$ in terms of $d\mu$ (understanding implicitly that $z \gt 0$ ) rather than $d\lambda$ :

f_{Z} (z) d z = g_{Z} (z) d μ = \frac{1}{2} {\begin{cases} 1 d z = z d μ, & if 0 \leq z \leq 1 \\ \frac{1}{z^{2}} d z = \frac{1}{z} d μ, & if z > 1. \end{cases}

$f_Z(z)\,dz = g_Z(z)\,d\mu = \frac{1}{2}\begin{cases} 1\,dz = z\, d\mu, & \text{if } 0 \le z \le 1 \\ \frac{1}{z^2}dz = \frac{1}{z}\, d\mu, & \text{if } z > 1. \end{cases}$

That is, the PDF with respect to the invariant measure $d\mu$ is $g_Z(z)$ , proportional to $z$ when $0\lt z \le 1$ and to $1/z$ when $1 \le z$ , close to what you had hoped.

This is not a mere one-off trick. Understanding the role of $d\mu$ makes many formulas look simpler and more natural. For instance, the probability element of the Gamma function with parameter $k$ , $x^{k-1}e^x\,dx$ becomes $x^k e^x d\mu$ . It's easier to work with $d\mu$ than with $d\lambda$ when transforming $x$ by rescaling, taking powers, or exponentiating.

The idea of an invariant measure on a group is far more general, too, and has applications in that area of statistics where problems exhibit some invariance under groups of transformations (such as changes of units of measure, rotations in higher dimensions, and so on).

— whuber
source

3

Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later.

— qed

4

If you think geometrically...

In the $X$ - $Y$ plane, curves of constant $Z = Y/X$ are lines through the origin. ( $Y/X$ is the slope.) One can read off the value of $Z$ from a line through the origin by finding its intersection with the line $X=1$ . (If you've ever studied projective space: here $X$ is the homogenizing variable, so looking at values on the slice $X=1$ is a relatively natural thing to do.)

Consider a small interval of $Z$ s, $(a,b)$ . This interval can also be discussed on the line $X=1$ as the line segment from $(1,a)$ to $(1,b)$ . The set of lines through the origin passing through this interval forms a solid triangle in the square $(X,Y) \in U = [0,1]\times[0,1]$ , which is the region we're actually interested in. If $0 \leq a < b \leq 1$ , then the area of the triangle is $\frac{1}{2}(1-0)(b-a)$ , so keeping the length of the interval constant and sliding it up and down the line $X=1$ (but not past $0$ or $1$ ), the area is the same, so the probability of picking an $(X,Y)$ in the triangle is constant, so the probability of picking a $Z$ in the interval is constant.

However, for $b>1$ , the boundary of the region $U$ turns away from the line $X = 1$ and the triangle is truncated. If $1 \leq a < b$ , the projections down lines through the origin from $(1,a)$ and $(1,b)$ to the upper boundary of $U$ are to the points $(1/a,1)$ and $(1/b,1)$ . The resulting area of the triangle is $\frac{1}{2}(\frac{1}{a} - \frac{1}{b})(1-0)$ . From this we see the area is not uniform and as we slide $(a,b)$ further and further to the right, the probability of selecting a point in the triangle decreases to zero.

Then the same algebra demonstrated in other answers finishes the problem. In particular, returning to the OP's last question, $f_Z(1/2)$ corresponds to a line that reaches $X=1$ , but $f_Z(2)$ does not, so the desired symmetry does not hold.

— Eric Towers
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3

Just for the record, my intuition was totally wrong. We are talking about density, not probability. The right logic is to check that

\int_{1}^{k} f_{Z} (z) d z = \int_{1 / k}^{1} f_{Z} (z) = \frac{1}{2} (1 - \frac{1}{k})

$\int_1^k f_Z(z) dz = \int_{1/k}^1 f_Z(z) = \frac{1}{2}(1 - \frac{1}{k})$ ,

and this is indeed the case.

— qed
source

1

Yea the link Distribution of a ratio of uniforms: What is wrong? provides CDF of $Z=Y/X$ . The PDF here is just derivative of the CDF. So the formula is correct. I think your problem lies in the assumption that you think Z is "symmetric" around 1. However this is not true. Intuitively Z should be a skewed distribution, for example it is useful to think when Y is a fixed number between $(0,1)$ and X is a number close to 0, thus the ratio would be going to infinity. So the symmetry of distribution is not true. I hope this help a bit.

— lzstat
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