两个独立的均匀随机变量之间的比率分布

Supppse $X$和$Y$在标准均匀分布在$[0, 1]$，它们是独立的，是什么的PDF $Z = Y / X$？

一些概率论教科书的答案是

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

我想知道，通过对称性，不应该$f_Z(1/2) = f_Z(2)$？根据上述PDF，情况并非如此。

右边的逻辑是，具有独立， ž = $X, Y \sim U(0,1)$和 Z−1=$Z=\frac YX$具有相同的分布，因此对于0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$ ，其中用的CDF式使用的事实，

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ 是一个连续随机变量，所以P{ž≥一个}=P{ž>一个}=1-˚FŽ（一）。因此Z的pdf满足 fZ（z）=z−2fZ（z−1）$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$ 因此 f Z（ $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ ，而不是fZ（$f_Z(\frac 12) = 4f_Z(2)$ 如您所想。$f_Z(\frac 12) = f_Z(2)$

这种分布是对称的-如果您以正确的方式看待它。

您（正确）观察到的对称性是和X / Y = 1 /（Y / X ）必须相同分布。使用比率和幂时，实际上是在正实数的乘法组中。位置不变测度的模拟d λ = d X上的添加剂的实数- [R是尺度不变测度d μ = d X /$Y/X$$X/Y = 1/(Y/X)$$d\lambda=dx$$\mathbb{R}$ $d\mu = dx/x$在正实数的乘法群上。它具有以下理想的属性：$\mathbb{R}^{*}$

1. 是不变的下变换 X → 一个X为任何正的常数一个： d μ （一个X ）= d （一个X $d\mu$$x\to ax$$a$

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

2. 就是协变下变换 X → X b为非零数 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. 被变换成 d λ经由指数： d μ （Ë X）= d È $d\mu$$d\lambda$ 同样，dλ变换回dμ经由对数。

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

（3）建立了测量组之间的同构和（- [R *，* ，d μ ）。反射X → - X上的添加剂空间对应于反转X → 1 / X上的乘法的空间，因为ë - X = 1 / Ë X。$(\mathbb{R}, +, d\lambda)$$(\mathbb{R}^{*}, *, d\mu)$$x \to -x$$x \to 1/x$$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.