두 개의 독립적 인 균일 랜덤 변수 사이의 비율 분포

Supppse $X$ 와 $Y$ 는 에서 균일하게 분포 된 표준 $[0, 1]$이며 독립적입니다. 의 PDF는 무엇 $Z = Y / X$입니까?

확률 이론 교과서의 대답은

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

I는 대칭으로 안 궁금 ? 위의 PDF에 따르면 그렇지 않습니다.$f_Z(1/2) = f_Z(2)$

올바른 논리는 독립 , Z = $X, Y \sim U(0,1)$ 및 Z-1=$Z=\frac YX$ 는 같은분포를 가지므로0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$ CDF를 사용한 방정식은

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ 는 연속 랜덤 변수이므로P{Z≥a}=P{Z>a}=1-FZ(a). 따라서 PDF의Z를만족 FZ(Z)=Z-2FZ(Z-1)$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$ 따라서 f Z ( $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ 및fZ(1아님$f_Z(\frac 12) = 4f_Z(2)$ 생각한대로 2 )=fZ(2).$f_Z(\frac 12) = f_Z(2)$

이 분포 는 대칭입니다. 올바른 방향으로 보면됩니다.

(정확하게) 관찰 한 대칭은 와 X / Y = 1 / ( Y / X ) 가 동일하게 분포되어야한다는 것입니다. 비율과 거듭 제곱으로 작업 할 때 실제로 양의 실수 의 곱셈 그룹 내에서 작업하고 있습니다 . 위치 측정 불변의 아날로그 D λ = D (X) 상의 첨가제 실제 숫자 R이 인 스케일 불변 측정 D μ = D (X) /$Y/X$$X/Y = 1/(Y/X)$$d\lambda=dx$$\mathbb{R}$ $d\mu = dx/x$양의 실수 의 곱셈 그룹 에서. 다음과 같은 바람직한 특성이 있습니다.$\mathbb{R}^{*}$

1. 는양의 상수 a : d μ ( a x ) = d ( a x ) 에 대해 변환 x → a x 에서 변하지 않습니다.$d\mu$$x\to ax$$a$

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

2. 는0이 아닌 숫자에 대한변환 x → x b 에서공변량입니다. 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. 는지수를 통해 d λ 로 변환됩니다. d μ ( e x ) = d e $d\mu$$d\lambda$ 마찬가지로,dλ는로그를 통해다시dμ로변환됩니다.

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

(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 $e^{-x} = 1/e^x$.

$Z=Y/X$$d\mu$$z \gt 0$$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.