दो स्वतंत्र वर्दी यादृच्छिक चर के बीच अनुपात का वितरण

Question

मान लीजिए कि $X$ और $Y$ मानक समान रूप से [ 0 , 1 ] में वितरित किए गए हैं $[0, 1]$, और वे स्वतंत्र हैं, Z = Y / X का PDF क्या है $Z = Y / X$?

कुछ संभाव्यता सिद्धांत पाठ्यपुस्तक का उत्तर है

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

मैं सोच रहा हूँ, समरूपता से, नहीं होना चाहिए $f_Z(1/2) = f_Z(2)$ ? ऊपर पीडीएफ के अनुसार ऐसा नहीं है।

Comments

$X$ और Y का डोमेन क्या है $Y$?

— सोबी

en.wikipedia.org/wiki/Ratio_distribution

— kjetil b halvorsen

आप यह सच होने की उम्मीद क्यों करेंगे? घनत्व समारोह आपको बताता है कि कसकर पैक संभावना एक बिंदु के पड़ोस में है, और यह स्पष्ट रूप से अधिक कठिन है के लिए $Z$ के पास होने के लिए $2$ से $1/2$ (उदाहरण है कि के लिए विचार $Z$ हमेशा हो सकता है $1/2$ कोई बात नहीं क्या $X$ है, लेकिन $Z < 2$ जब $X > 1/2$ )।

— dsaxton

वर्दी के अनुपात

— शीआन

मुझे नहीं लगता कि यह एक डुप्लिकेट है, यह प्रश्न पीडीएफ की मांग कर रहा है, यहां मेरे पास पीडीएफ है, मैं सिर्फ इसकी शुद्धता (शायद भोलेपन) पर सवाल उठा रहा हूं।

— QED

Answer 1

सही तर्क है कि स्वतंत्र के साथ एक्स , वाई ~ यू ( 0 , 1 ) , जेड = $X, Y \sim U(0,1)$एक्स और जेड-1=$Z=\frac YX$Y का समानवितरण हैऔर इसलिए0<z<1 P { $Z^{-1} =\frac XY$$0 < z < 1$ X ≤z}= पी { एक्सY ≤z}= पी { यएक्स ≥1z }FZ(z)= 1 - एफ जेड ( 1)z ) जहां CDF के साथ समीकरण इस तथ्य का उपयोग करता है कि

$$\begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align}$$ एक्स एक सतत यादृच्छिक चर है और इसलिएपी{जेड≥एक}=पी{जेड>एक}=1-एफजेड(एक)। इसलिए की पीडीएफजेडसंतुष्ट चजेड(जेड)=z-2चजेड(जेड-1$\frac YX$$P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$$Z$ ,0 < z < 1. इस प्रकार एफ जेड ( $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ 2 )=4एफजेड(2), औरएफजेड(1)नहीं$f_Z(\frac 12) = 4f_Z(2)$2 )=fZ(2) जैसा आपने सोचा कि यह होना चाहिए।$f_Z(\frac 12) = f_Z(2)$

Answer 2

यह वितरण है सममित - यदि आप इसे पर सही तरीके से लग रहे हो।

आपके पास जो समरूपता है (सही ढंग से) देखी गई है कि Y / X और X / Y = 1 / ( Y / X ) को समान रूप से वितरित किया जाना चाहिए। अनुपात और शक्तियों के साथ काम करते समय, आप वास्तव में सकारात्मक वास्तविक संख्याओं के गुणक समूह के भीतर काम कर रहे हैं । स्थान अपरिवर्तनीय उपाय के अनुरूप घ λ = घ एक्स पर additive वास्तविक संख्या आर है अपरिवर्तनीय पैमाना उपाय घ μ = घ एक्स /$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.

---

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).

Comments

Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later.

— qed

Answer 3

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.

Answer 4

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.

Answer 5

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.