Independant Bernoulli Random Varible

**wrexlive1** · February 21st 2013, 03:28 AM

SupposeY1; Y2; Y3; : : : are independent Bernoulli random variables with

P(Yi = 0) = 1 =(1/i!)
P(Yi = 1) = 1=i!

Defining X=Y1+Y2+Y3+...Show that E[x]=e-1

Not to sure how to do this queston, so wondering anyone can help me?

**Siron** · February 21st 2013, 09:00 AM

Note that the expected value $E$ has linear properties which we can use here. Unfortunately I'm confused with your notations. Can you explain what you mean with $1=\frac{1}{i!}$ ? We know that $Y_1,Y_2, \ldots$ are discrete randiom variables, hence $E[Y_i] = \sum_{x} x p_x$ . Finally, $E[X] = \sum_{i=1}^{\infty} E[Y_i]$

**wrexlive1** · February 21st 2013, 09:25 AM

Sorry there has been a typing error, it should read as
P(Yi = 0) = 1 -(1/i!)
P(Yi = 1) = 1/i!

**Siron** · February 21st 2013, 09:38 AM

Originally Posted by wrexlive1

Sorry there has been a typing error, it should read as
P(Yi = 0) = 1 -(1/i!)
P(Yi = 1) = 1/i!

Okay. We can compute the expected value of $Y_i$ as follows $E[Y_i] = 0\left(1-\frac{1}{i!}\right)+1\left(\frac{1}{i!}\right)$ $=\frac{1}{i!}$ .
Therefore, $E[X]=\sum_{i=1}^{\infty} E[Y_i]=\sum_{i=1}^{\infty} \frac{1}{i!} = \sum_{i=0}^{\infty} \frac{1}{i!}-1$ . Using the fact that $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$ we have $\sum_{i=0}^{\infty} \frac{1}{i!}=e$ . Hence, $E[X]=e-1$ .

Thread: Independant Bernoulli Random Varible

LinkBack

Thread Tools

Display

Independant Bernoulli Random Varible

Re: Independant Bernoulli Random Varible

Re: Independant Bernoulli Random Varible

Re: Independant Bernoulli Random Varible

Similar Math Help Forum Discussions

Two Bernoulli Random Variables

Bernoulli Random Variable

independant random variables ?

Probability and random random varible question

Combinations of independant normal random variables

Search Tags