दो स्वतंत्र वर्दी यादृच्छिक चर के बीच अनुपात का वितरण


17

मान लीजिए कि XX और YY 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

जेड ( जेड ) = { 1 / 2 , अगर  0 जेड 1 1 / ( 2 जेड 2 ) , यदि  z > 1 0 , अन्यथा

fZ(z)=1/2,1/(2z2),0,if 0z1if z>1otherwise.

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


XX और Y का डोमेन क्या है Y?
सोबी


2
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 जेडZ के पास होने के लिए 22 से 1 / 21/2 (उदाहरण है कि के लिए विचार जेडZ हमेशा हो सकता है 1 / 21/2 कोई बात नहीं क्या एक्सX है, लेकिन जेड < 2Z<2 जब एक्स > 1 / 2X>1/2 )।
dsaxton


3
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

जवाबों:


19

सही तर्क है कि स्वतंत्र के साथ एक्स , वाई ~ यू ( 0 , 1 ) , जेड = YX,YU(0,1)एक्स और जेड-1=एक्सZ=YXY का समानवितरण हैऔर इसलिए0<z<1 P { Y के लिए हैZ1=XY0<z<1 Xz}= पी { एक्सYz}= पी { एक्स1z }FZ(z)= 1 - एफ जेड ( 1)z ) जहां CDF के साथ समीकरण इस तथ्य का उपयोग करता है किY

P{YXz}FZ(z)=P{XYz}=P{YX1z}=1FZ(1z)
एक्स एक सतत यादृच्छिक चर है और इसलिएपी{जेडएक}=पी{जेड>एक}=1-एफजेड(एक)। इसलिए की पीडीएफजेडसंतुष्ट जेड(जेड)=z-2जेड(जेड-1)YXP{Za}=P{Z>a}=1FZ(a)Z ,0 < z < 1. इस प्रकार एफ जेड ( 1)
fZ(z)=z2fZ(z1),0<z<1.
2 )=4एफजेड(2), औरएफजेड(1)नहींfZ(12)=4fZ(2)2 )=fZ(2) जैसा आपने सोचा कि यह होना चाहिए।fZ(12)=fZ(2)

14

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 λ = एक्स पर additive वास्तविक संख्या आर है अपरिवर्तनीय पैमाना उपाय μ = एक्स / एक्सY/XX/Y=1/(Y/X)dλ=dxR dμ=dx/x on the multiplicative group RR of positive real numbers. It has these desirable properties:

  1. dμdμ is invariant under the transformation xaxxax for any positive constant aa: dμ(ax)=d(ax)ax=dxx=dμ.

    dμ(ax)=d(ax)ax=dxx=dμ.
  2. dμdμ is covariant under the transformation xxbxxb for nonzero numbers bb: dμ(xb)=d(xb)xb=bxb1dxxb=bdxx=bdμ.

    dμ(xb)=d(xb)xb=bxb1dxxb=bdxx=bdμ.
  3. dμdμ is transformed into dλdλ via the exponential: dμ(ex)=dexex=exdxex=dx=dλ.

    dμ(ex)=dexex=exdxex=dx=dλ.
    Likewise, dλdλ is transformed back to dμdμ via the logarithm.

(3) establishes an isomorphism between the measured groups (R,+,dλ)(R,+,dλ) and (R,,dμ)(R,,dμ). The reflection xxxx on the additive space corresponds to the inversion x1/xx1/x on the multiplicative space, because ex=1/exex=1/ex.

Let's apply these observations by writing the probability element of Z=Y/XZ=Y/X in terms of dμdμ (understanding implicitly that z>0z>0) rather than dλdλ:

fZ(z)dz=gZ(z)dμ=12{1dz=zdμ,if 0z11z2dz=1zdμ,if z>1.

fZ(z)dz=gZ(z)dμ=12{1dz=zdμ,1z2dz=1zdμ,if 0z1if z>1.

That is, the PDF with respect to the invariant measure dμdμ is gZ(z)gZ(z), proportional to zz when 0<z10<z1 and to 1/z1/z when 1z1z, close to what you had hoped.


This is not a mere one-off trick. Understanding the role of dμdμ makes many formulas look simpler and more natural. For instance, the probability element of the Gamma function with parameter kk, xk1exdxxk1exdx becomes xkexdμxkexdμ. It's easier to work with dμdμ than with dλdλ when transforming xx 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).


3
Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later.
qed

4

If you think geometrically...

In the XX-YY plane, curves of constant Z=Y/XZ=Y/X are lines through the origin. (Y/XY/X is the slope.) One can read off the value of ZZ from a line through the origin by finding its intersection with the line X=1X=1. (If you've ever studied projective space: here XX is the homogenizing variable, so looking at values on the slice X=1X=1 is a relatively natural thing to do.)

Consider a small interval of ZZs, (a,b)(a,b). This interval can also be discussed on the line X=1X=1 as the line segment from (1,a)(1,a) to (1,b)(1,b). The set of lines through the origin passing through this interval forms a solid triangle in the square (X,Y)U=[0,1]×[0,1](X,Y)U=[0,1]×[0,1], which is the region we're actually interested in. If 0a<b10a<b1, then the area of the triangle is 12(10)(ba)12(10)(ba), so keeping the length of the interval constant and sliding it up and down the line X=1X=1 (but not past 00 or 11), the area is the same, so the probability of picking an (X,Y)(X,Y) in the triangle is constant, so the probability of picking a ZZ in the interval is constant.

However, for b>1b>1, the boundary of the region UU turns away from the line X=1X=1 and the triangle is truncated. If 1a<b1a<b, the projections down lines through the origin from (1,a)(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 12(1a1b)(10). 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, fZ(1/2) corresponds to a line that reaches X=1, but fZ(2) does not, so the desired symmetry does not hold.


3

Just for the record, my intuition was totally wrong. We are talking about density, not probability. The right logic is to check that

k1fZ(z)dz=11/kfZ(z)=12(11k)

,

and this is indeed the case.


1

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.

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