5

How do I find the distribution of the weighted sum of independent Bernoulli random variables if the weights are non-negative real numbers?

I have N number of independent Bernoulli distributed random variables lets say $X_1, X_2, X_3...X_N$ and suppose I have a set of weights $W_1, W_2, W_3,...,W_N$ which are non-negative real numbers and the sum of all the weights is equal to $N$.

Then I need to find the distribution of the random variable $Z$ which is the weighted sum of the $N$ independent Bernoulli random variables i.e, $Z =\sum_i^N W_i*X_i$, where $i=1,2,\cdots,N$.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
nuria
  • 51
  • 4
  • This is conditional on the $W$'s? Do the $X_i$ all share the same $p$? – Glen_b Aug 18 '14 at 02:01
  • No, all Xi have different probability P and I am assuming the weight parameter W is constant for all Xi`s it doesn`t change. If Xi is highly probable then corresponding weight Wi will take large value between 0 and N. Thanks inadvance, Glen_b – nuria Aug 20 '14 at 06:52
  • Should answers be exact or approximate? – Glen_b Aug 20 '14 at 08:04
  • should be exact – nuria Aug 21 '14 at 01:22
  • 1
    Okay, but I don't think you'll much like the answer. – Glen_b Aug 21 '14 at 02:00
  • Will not give you exact answers, but might be very close: saddlepoint approximation. See https://stats.stackexchange.com/questions/191492/how-does-saddlepoint-approximation-work/191781#191781 and search this site. – kjetil b halvorsen Apr 14 '18 at 13:05
  • @Glen_b if they all shared the same $p$ and it didn't matter if it was approximate or exact would there be an answer? The weights in my situation may be dependent but the Bernoulli random variables are independent – RAND Oct 25 '18 at 05:47
  • 1
    When you say the weights are dependent, you're indicating that the *weights* are random variables rather than constants, unlike the situation here? – Glen_b Oct 25 '18 at 05:58

0 Answers0