OOps! I see I misread the problem. Since X = e_0 + X/2 X = 2e_0 and P(X=2)=1/2 P(X=-2)=1/2 Brent On 2/13/2014 1:42 PM, meekerdb wrote:
It's -inf with probability 1/2 and +inf with probability 1/2.
Brent
On 2/13/2014 11:18 AM, Dan Asimov wrote:
Let e_j, j = 1,2,3,… be independent random variables each taking the values +-1 with probability 1/2.
Let the random variable X be defined as
X := Sum_{n=1…oo} e_j/2^j.
PUZZLE: What is the distribution of X ???
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