# Random variable

1. Oct 9, 2005

### playboy

Hello...

hmm.. i am working on a homework problem, and im kina stuck.

the question reads: Suppose that X is a random variable which can take on any non-negative integer (including 0). Write P(X greater than and equal to i) in terms of the probability mass function of X and hence show that

E[X] = the sum of infinity, i = 1 P(X greater than and equal to i)

I tried to solve this problem by just exanding it i times.

For example, i suppose i = 0, 1, 2, 3, 4 .....

So the probability mass funtion would look like:

P(1) = P{X = 1}
P(2) = P{X = 2}
P(3) = P{X = 3}
P(4) = P{X = 4}

i times.. etc.

but getting E[X] has got be completely lost

I thought perhaps that E[X] = 1P(X = 1) + 2P(X = 2) + 3P(X = 3) ..... but what are the values of the mass function?

Anybody have an idea?

2. Oct 9, 2005

### playboy

Anyways, i got MUCH further than before, but still not quite their.

P(X >/= i) = 1 - P(X=0) - P(X=1) - ... - P(X=(i-1)) for i = 0,1,2,3,...

and so E[x] = -1 - 0(P(X=0)) - 1(P(X=1)) - ..... - (i-1)P(X=(i-1))

and then the sum of infinity at n = 1 is p(i)P[X>/=i]

This is where i get stuck.. i dont know how to convert/show "the sum of
infinity at n = 1 is p(i)P[X/>=i] " is equal to "the sum of infinity at
n = 1 is P[X>/=i] "

I know the above is all messy... and I think I almost on the got it..
but can anybody help me out?