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Probability expectations with n

Carla1985

Member
Feb 14, 2013
93
I'm stuck on this question. I've done plenty of the questions with numbers etc but not sure how to deal with the n case for this one.

Screen Shot 2013-03-10 at 21.40.25.png

Thank you :)
 

chisigma

Well-known member
Feb 13, 2012
1,704
I'm stuck on this question. I've done plenty of the questions with numbers etc but not sure how to deal with the n case for this one.

View attachment 689

Thank you :)
If the discrete p.d.f. is...

$\displaystyle P \{ \xi = n \} = \frac{3}{4}\ (\frac{1}{4})^{n}$ (1)

... then, tacking into account that is...

$\displaystyle \sum_{n=0}^{\infty} n\ x^{n} = \frac{x} {(1-x)^{2}}$ (2)

... You obtain...

$\displaystyle E \{ \xi \} = \frac{3}{4}\ \sum_{n=0}^{\infty} n\ (\frac{1}{4})^{n} = \frac{\frac{3}{4}\ \frac{1}{4}}{(\frac{3}{4})^{2}} = \frac{1}{3}$ (3)

Regarding (ii) by definition is...

$\displaystyle E \{ (-3)^{\xi} \} = \frac{3}{4}\ \sum_{n=0}^{\infty} (-3)^{n}\ (\frac{1}{4})^{n} = \frac{3}{4} \frac{1}{1+\frac{3}{4}} = \frac{3}{7}$ (4)


Kind regards


$\chi$ $\sigma$
 
Last edited:

Carla1985

Member
Feb 14, 2013
93
... You obtain...

$\displaystyle E \{ \xi \} = \frac{3}{4}\ \sum_{n=0}^{\infty} n\ (\frac{1}{4})^{n} = \frac{\frac{3}{4}}{(\frac{3}{4})^{2}} = \frac{4}{3}$ (3)

Im confused by this part. If i use the series from the line above taking x to be 1/4 isnt it 1/4 on top instead of 3/4? or have i missed something?
Thanks for the help x
 

chisigma

Well-known member
Feb 13, 2012
1,704
Im confused by this part. If i use the series from the line above taking x to be 1/4 isnt it 1/4 on top instead of 3/4? or have i missed something?
Thanks for the help x
All right!... the mistake has been corrected... thak You very much!...

Kind regards

$\chi$ $\sigma$
 

Carla1985

Member
Feb 14, 2013
93
Thats fab, thanks very much for the help x