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Expected value and equality to sums

How to show that$E[N]=\displaystyle\sum_{k=1}^\infty P{\{N\geq k\}}=\displaystyle\sum_{k=0}^\infty P{\{N>k\}}$

If any member here knows the answer, may reply to this question.:confused:
 
How to show that$E[N]=\displaystyle\sum_{k=1}^\infty P{\{N\geq k\}}=\displaystyle\sum_{k=0}^\infty P{\{N>k\}}$

If any member here knows the answer, may reply to this question.:confused:
Hello,
'N' denote a non-negative integervalued random variable.
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,648
St. Augustine, FL.
Hello,

I got the answer after doing some carefully thinking.
Perhaps yu'd like to share your solution so that others facing the same or similar question can benefit from your work?
 
Hello,
If we define the sequence of random variable $I_n$ (Indicator random variable), n > 1 by

$$I_n= \left \{ {1,\text{if n < X} \atop \text{0, if n>X}} \right.$$. Now express X in terms of $I_n.$ (Actually, I don't know how to express in terms of $I_n$:confused:)

I understood the equation in #1 by using the expectation of random variable X(outcome of a toss of a fair dice)is equal to summation of the probabilities of X > n, where range of n is 0 to $\infty$

I think the following below mentioned identities will be useful here.

$$ a)(1-1)^N= \left \{{\text{1, if N > 0}\atop \text{0, if n < 0}} \right.$$
$$b)(1-1)^N=\displaystyle\sum_{n=0}^n\binom{N}{i}*(-1)^i$$

$$ c)1-I=\displaystyle\sum_{n=0}^n\binom{N}{i}*(-1)^i$$

$$ d)I=\displaystyle\sum_{n=1}^n\binom{N}{i}*(-1)^i$$

If you want to show this equation in mathematical language, you may reply to that effect.:)
 
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