Distribución de la relación entre dos variables aleatorias uniformes independientes

Supppse $X$ e $Y$ están distribuidos uniformemente de manera uniforme en $[0, 1]$ , y son independientes, ¿cuál es el PDF de $Z = Y / X$ ?

La respuesta de algún libro de texto de teoría de la probabilidad es

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

Me pregunto, por simetría, no debe $f_Z(1/2) = f_Z(2)$ ? Este no es el caso según el PDF anterior.

La lógica correcta es que con X , Y ∼ U ( 0 , 1 ) independientes , Z = $X, Y \sim U(0,1)$X y Z-1=$Z=\frac YX$Y tiene la mismadistribucióny por lo tanto para0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ donde la ecuación con CDF utiliza el hecho de que YX es una variable aleatoria continua y, por lo tanto,P{Z≥a}=P{Z>a}=1-FZ(a). Por lo tanto, el pdf deZsatisface fZ(z)=z-2fZ(z-1)$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$0 < z < 1. Así $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ f Z ( 12 )=4fZ(2), y no$f_Z(\frac 12) = 4f_Z(2)$ Z ( 12 )=fZ( 2 ) como pensabas que debería ser.$f_Z(\frac 12) = f_Z(2)$

Esta distribución es simétrica, si la miras de la manera correcta.

La simetría que ha observado (correctamente) es que Y / X y X / Y = 1 / ( Y / X ) deben distribuirse de manera idéntica. Cuando trabajas con razones y potencias, realmente estás trabajando dentro del grupo multiplicativo de los números reales positivos. El análogo de la medida invariante de ubicación d λ = d x en los números reales aditivos R es la medida invariante de escala d μ = d x /$Y/X$$X/Y = 1/(Y/X)$$d\lambda=dx$$\mathbb{R}$ $d\mu = dx/x$ en el grupo multiplicativo $\mathbb{R}^{*}$ of positive real numbers. It has these desirable properties:

1. $d\mu$ is invariant under the transformation $x\to ax$ for any positive constant $a$:

   $$d\mu(ax) = \frac{d(ax)}{ax} = \frac{dx}{x} = d\mu.$$

2. $d\mu$ is covariant under the transformation $x\to x^b$ for nonzero numbers $b$:

   $$d\mu(x^b) = \frac{d(x^b)}{x^b} = \frac{b x^{b-1} dx}{x^b} = b\frac{dx}{x} = b\, d\mu.$$

3. $d\mu$ is transformed into $d\lambda$ via the exponential:

   $$d\mu(e^x) = \frac{de^x}{e^x} = \frac{e^x dx}{e^x} = dx = d\lambda.$$ 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 $d\lambda$:

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

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.

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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).

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.

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

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

and this is indeed the case.

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.