Question

2.42. Chebyshev inequality. Show that the Chebyshev inequality, Eq. (2.4.11) holds by: (a) Splitting the integral defining o2 into three intervals, — o to m— ho, m— ho to m + ho, and m+ ho to co. (b) Showing that ch 022/". * oh=f(x) dx + Samotn oh=f(e) de 2 oh?(P[X <m - h] + P[X > m + h]) and hence that Eq. (2.4.11) holds. If a random variable has known m equal to 5000 and known o equal to (1000)- but with unknown distribution (owing, say, to the intractable mathematics involved in its derivation), find the ranges within which the variable will lie with probabilities at least 0.5, 0.75, 0.90, and 0.99.

Answer 1

Ans! chebyshev nequalty a andom varable x usth Suppose we have Pho babtity functon assume that e[x2 < oo . Then and h>o. we have EG) =m mch mth -m)fa)dz+@-m) f&dx« (&m)f@sdx- mth mth mh (2-m)d m5f@d1 &m) faduzo mth mth m-h I M-h and 12 mth Snce e, pesitve -mzh mth F&)dzt f)di (P(KE m-h) mt

m-hand x > mth) 2 x-mizh) x-m h I equation G) hs replace by hthen, we have mh 2 eix-ml ch P-2h C-P-mt2he2 -4 from2 2 Pm-hx <m thJ21-

Preblem Yandom vartable wth Ex) =m =So00. Let x be a varx) 2 (lo00) and o00 e Cix-ml h20.5 PSooo-h l000 x 5000hx1000 equations and Fom ,we get hen sooo - ul xloo0 <x 2 s000 +uIu xlo002 0.S GHI4205 Kange of 1S 3se6, Gulu) f PIx-mlho205 Then, from we get 1 - - 3 m-h mt h 2045 X

4 Pso00-2xlo00 < x 5000 2x 1000 20.7S X < 1000, 3000 from we get hen Pm-hxmt hes J 2040. e sooo -3 x to00 c x < scoo t3x0o02040 PC2000 x 800] 2090. 2000x 2000. Cv) I P I-m eh20-9 Then, fom we get l00 cx E m th%J 2 099 -lo X laooX So00 + 10Xio0» 099. 62 e Lm-h seco p-soco x ISoo0 20I9. -SOoo X < IS000