Question

Can anyone help me understand this? Let X be a random variable that equals k>0 with probability 1/(k^2+1); otherwise, X = 0. Compute E[X].

Answer 1

i understand expected values but im unsure how i would do this problem

Answer 2

if you know the formula for expected value can you tell me?

Answer 3

ya so i think it would be the integral from 0 to infinity of k(1/(k^2+1))

Answer 4

becasue usually say you wanted to find the E[S] and you have f(S) it is just the sum of sf(s)

Answer 5

great!!
so now k/(k^2 + 1) can be written as (1/2) * (2k/ (k^2 +1))
can you now figure out something?

Answer 6

would that just be u substituion?

Answer 7

so (1/2) ln(k^2+1) from 0 to infinity?

Answer 8

that cant be right because i vant have an expected value that does not converge can i?

Answer 9

i think it will because as we move towards larger values it`s probability decreases and so does p*x

Answer 10

ok so the next part asks for the variance. how could i compute the variance if E[x]^2 would be infinity^2?

Answer 11

variance is integral k^2/ (k^2 +1)
integral of 1 - 1/(k^2 +1)
= k - arctan(k) {limits 0 to infinity}

Answer 12

right so varience is E[x^2] (which is what you wrote) - E[x]^2 (what the other part is squared)?

Answer 13

(E[x]) ^ 2 is what you are asking for ?

Answer 14

no im saying if we computed E[x] to be infinity in the previous part. then when finding E[x}^2 we would get infinity^2 which seems strange to me

Answer 15

actually i have a feeling that we have screwed up somewhere logically this expected value should not approach infinity due to the reasons for larger values their probability of occurrence is very less

i feel it would be better to take advice of someone

Answer 16

haha its not a problem thank you for your help though

Answer 17

well the integral is from -infinity to infinity in this case not from 1

Answer 18

so you think maybe it is just 0? that would still be 0-inf^2

Answer 19

in such integrals we use something like
given that k/(k^2 - 1) is an odd function for the values for x and -x are exactly opposite of each other and they cancel out so this integral must come out to be zero

Answer 20

oh holdon i just got an email from my professor. another student said " In class, we discussed that the givens in problem 4 provide us with
> information to say that X = kY, where Y is Bernoulli(1/(k^2 + 1)).
" and my professor responded with "It does; however, it would have been better if I had written that X is
either k + 1/k with probability 1/(k^2 + 1) since this would help you
out with part d). "

Answer 21

"you may either
use either the p.m.f. that I gave in the question or you may use the
p.m.f. where X is k + 1/k with probability k^2/(k^2 + 1) and otherwise
zero. "