Question

Suppose Y~p(y)=1/30(y-5)^2 for y = 1,2,3,4. Calculate E(1/Y).

Answer 1

where are you stuck?

Answer 2

I don't even know how to start.

Answer 3

do you know how to find expected values?

Answer 4

the summation of the y*p(y)

Answer 5

\[E(G(Y))=\sum_{y\in\Omega}G(y)P(Y=y)\]

Answer 6

here your \(G(y)=\frac{1}{y}\)

Answer 7

so the summation should be: \[\sum_{?}^{?}\frac{ 1 }{ y }(\frac{ 1 }{ 30 })(y-5)^2 \] ?

Answer 8

\[\sum_{y=1}^{4}\frac{ 1 }{ y }\left(\frac{ 1 }{ 30 }\right)(y-5)^2\]

Answer 9

right, okay. my prof also gave me r=1-p. Is that going to be useful at all? or is it just extra information?

Answer 10

I would just compute the above

Answer 11

okay. Thank you.

Answer 12

I get 53/72

Answer 13

I got the same. Thank you again!