- Thread starter
- #1

#### Dhamnekar Winod

##### Active member

- Nov 17, 2018

- 100

If any member here knows the answer, may reply to this question.

- Thread starter Dhamnekar Winod
- Start date

- Thread starter
- #1

- Nov 17, 2018

- 100

If any member here knows the answer, may reply to this question.

- Thread starter
- #2

- Nov 17, 2018

- 100

Hello,

If any member here knows the answer, may reply to this question.

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

- Thread starter
- #3

- Nov 17, 2018

- 100

Hello,Hello,

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

I got the answer after doing some carefully thinking.

- Admin
- #4

Perhaps yu'd like to share your solution so that others facing the same or similar question can benefit from your work?Hello,

I got the answer after doing some carefully thinking.

- Thread starter
- #5

- Nov 17, 2018

- 100

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.

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.

Last edited: