Expected value and equality to sums

[WMDhamnekar]

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.

 

[WMDhamnekar]

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.

Hello,
'N' denote a non-negative integervalued random variable.

 

[WMDhamnekar]

Hello,
'N' denote a non-negative integervalued random variable.

Hello,

I got the answer after doing some carefully thinking.

 

[MarkFL]

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?

 

[WMDhamnekar]

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 $I_n$)

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.:)