Distribuzione del rapporto tra due variabili casuali uniformi indipendenti

Supponiamo che $X$ e $Y$ siano distribuiti uniformemente in $[0, 1]$ e sono indipendenti, qual è il PDF di $Z = Y / X$ ?

La risposta di un manuale di teoria delle probabilità è

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

Mi chiedo, per simmetria, non dovrei $f_Z(1/2) = f_Z(2)$ ? Questo non è il caso secondo il PDF sopra.

La logica giusta è quella con indipendente , Y ∼ U ( 0 , 1 ) , Z = $X, Y \sim U(0,1)$ e Z-1=$Z=\frac YX$ ha la stessadistribuzionee quindi per0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$ dove l'equazione con i CDF usa il fatto che

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ è una variabile casuale continua e quindiP{Z≥a}=P{Z>a}=1-FZ(a). Quindi il pdf diZsoddisfa fZ(z)=z-2fZ(z-1)$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$ Quindi f Z ( $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ e nonfZ($f_Z(\frac 12) = 4f_Z(2)$ come pensavi dovesse essere.$f_Z(\frac 12) = f_Z(2)$

Questa distribuzione è simmetrica, se la guardi nel modo giusto.

La simmetria che hai (correttamente) osservato è che e X / Y = 1 / ( Y / X ) devono essere distribuiti in modo identico. Quando lavori con rapporti e poteri, stai davvero lavorando all'interno del gruppo moltiplicativo dei numeri reali positivi. L'analogo della misura invariante di posizione d λ = d x sui numeri reali additivi R è la misura invariante di scala d μ = d x /$Y/X$$X/Y = 1/(Y/X)$$d\lambda=dx$$\mathbb{R}$ $d\mu = dx/x$sul gruppo moltiplicativo di numeri reali positivi. Ha queste proprietà desiderabili:$\mathbb{R}^{*}$

1. è invariante sotto la trasformazione x → a x per qualsiasi costante positiva a : d μ ( a x ) = d ( a x $d\mu$$x\to ax$$a$

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

2. è covariante sotto la trasformazione x → x b per numeri diversi da zero b : d μ ( x b ) = d ( x b$d\mu$$x\to x^b$$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. viene trasformato in d λ tramite l'esponenziale: d μ ( e x ) = d e $d\mu$$d\lambda$ Allo stesso modo,dλviene trasformato indμtramite il logaritmo.

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

(3) stabilisce un isomorfismo tra i gruppi misurati e ( R ∗ , ∗ , d μ ) . La riflessione x → - x sullo spazio additivo corrisponde all'inversione x → 1 / x sullo spazio moltiplicativo, perché e - x = 1 / e x .$(\mathbb{R}, +, d\lambda)$$(\mathbb{R}^{*}, *, d\mu)$$x \to -x$$x \to 1/x$$e^{-x} = 1/e^x$

$Z=Y/X$$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.

---

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.