Oranın iki bağımsız düzgün rasgele değişken arasında dağılımı

Supppse $X$ ve $Y$ standart olarak [ 0 , 1 ] 'de eşit olarak dağılmıştır $[0, 1]$ve bağımsızdırlar, Z = Y / X'in PDF dosyası $Z = Y / X$nedir?

Bazı olasılık teorisi ders kitabının cevabı

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

Merak ediyorum, simetri ile, ? Yukarıdaki PDF'ye göre durum böyle değil.$f_Z(1/2) = f_Z(2)$

Doğru mantık, bağımsız X , Y ∼ U ( 0 , 1 ) , Z = $X, Y \sim U(0,1)$X ve Z-1=$Z=\frac YX$Y aynıdağılıma sahiptirve bu nedenle0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$ X ≤z}= P { XY ≤Z}= P { YX ≥1z }FZ(z)= 1 - F Z ( 1z ) CDF'lerle denklemin

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ X sürekli rasgele bir değişkendir ve dolayısıylaP{Z≥a}=P{Z>a}=1-FZ(a). DolayısıylaZ'ninpdf'si fZ(z)=z-2fZ(z-1) 'i$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$0 < z < 1. Böylece f Z ( $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ 2 )=4fZ(2),fZ(1değil$f_Z(\frac 12) = 4f_Z(2)$2 )=fZ(2) olması gerektiği gibi.$f_Z(\frac 12) = f_Z(2)$

Bu dağıtım olan simetrik - Nereden bakarsanız doğru yolu bakarsan.

(Doğru) gözlemlediğiniz simetri, Y / X ve X / Y = 1 / ( Y / X ) 'nin aynı şekilde dağıtılması gerektiğidir. Oranlar ve güçlerle çalışırken, gerçekten pozitif gerçek sayıların çarpım grubu içinde çalışıyorsunuz . Yer değişmez ölçüsü analog d λ = d x üzerine ilave reel sayılar R olan ölçek değişmez ölçü d μ = d x /$Y/X$$X/Y = 1/(Y/X)$$d\lambda=dx$$\mathbb{R}$ $d\mu = dx/x$ on the multiplicative group $\mathbb{R}^{*}$ 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.