2018 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Now consider the following picture in which R represents the points, (||X || , ||Y ||) such that

1√2|||X ||− ||Y ||| ≤ s and

1√2(||X ||+ ||Y ||)> t.

(?, t−s√2)

( t−s√2,?)

R

Therefore, continuing with the chain of inequalities above,

P(||X || ≤ s)P(||Y ||> t)

≤ P(||X ||> t− s√

2, ||Y ||> t− s√

2

)= P

(||X ||> t− s√

2

)2

.

Since X ,Y have the same law, this can be written as

P(||X ||> t)≤P(||X ||> t−s√

2

)2

P(||X || ≤ s).

Now define a sequence as follows. t0 ≡ r > 0 and tn+1 ≡ r +√

2tn. Also, in the aboveinequality, let s≡ r and then it follows

P(||X ||> tn+1) ≤P(||X ||> tn+1−r√

2

)2

P(||X || ≤ r)

=P(||X ||> tn)

2

P(||X || ≤ r).

Let

αn (r)≡P(||X ||> tn)P(||X || ≤ r)

.

Then it follows

αn+1 (r)≤ αn (r)2 , α0 (r) =

P(||X ||> r)P(||X || ≤ r)

.