here - Sites personnels de TELECOM ParisTech - Télécom ParisTech

2.6. Asymptotic behavior: β i → ∞ 69Proposition 2.6.5. For d ≥ 1 and k ≥ 0,P [X d = k] = 1 1 ∑∞ ( )( ) d − 1 + j d − 1 + k + j 14 d 2 k d − 1 d − 1 4 jj=0= 1 ( )1 d − 1 + k4 d 2 k 2F 1 (d, d + k; k + 1; 1/4) ,d − 1so thatP d = P [X d = 0] = 1 4 d∞ ∑j=0( d − 1 + jd − 1) 214 j = 1 4 d 2 F 1 (d, d; 1; 1/4) .In particular 13 d ≤ P d ≤ 1+3·2d−24 d . Moreover, P 1 = 1/3 and P 2 = 5/27.Proof. To get the expression of P [X d = k] as a power series, the idea is to split it according tothe value of one of the two sums of d random variables (the value of the other sum is then alsofixed) and to use the above lemma:[∞∑ d∑] [ d∑]P [X d = k] = P G i = j P H i = j + kj=0 i=1i=1]]∞∑= P G i = j P H i = j + kj=0= 1 4 d 12 k ∞ ∑[ d∑[ d∑i=1i=1( )( ) d − 1 + j d − 1 + k + j 1d − 1 d − 1 4 j .This power series is easily seen to be equal to( )1 1 d − 1 + k4 d 2 k 2F 1 (d, d + k; k + 1; 1/4) .d − 1j=0Setting k = 0 in the above expressions givesP d = P [X d = 0] = 1 4 d∞ ∑j=0( d − 1 + jd − 1This power series can be bounded from below by1 ∑∞4 d j=0( d − 1 + jd − 1) 214 j = 1 4 d 2 F 1 (d, d; 1; 1/4) .) 14 j = 1 4 d 1(1 − 1/4) d = 1 3 d ,and from above by⎛⎞1∞∑( ) d − 1 + j 2d−2+j⎝14 d +⎠d − 1 4 j = 1 4 d + 2d−2 ∑∞ ( ) d − 1 + j 14 d d − 1 2 j − 2d−24 dj=1which gives the desired inequalities.j=0= 1 + 4d−1 − 2 d−24 d = 1 + 3 · 2d−24 d ,