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Expected Value of reciprocal (Sorry for reposting)

by giglamesh
Tags: expected, reciprocal, reposting
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giglamesh
#1
Nov4-11, 02:15 PM
P: 14
Hi all
Sorry for reposting, the previous post wasn't clear enough, it's my mistake, I'll make the question more clear, I found lot of people asking the same question on the Internet.

Given that X is random variable that takes values:

0..............H-1

The PMF of X is unknown, but I can tell what is the expected value which is [itex]\bar{X}[/itex]

There is event Y when calculated it gives the value:

[itex]P(Y)=E[\frac{1}{X+1}][/itex]

The QUESTION: Is there a way to find expected value [itex]\bar{Y}[/itex] in the terms of [itex]\bar{X}[/itex]? regarding that: the PMF of X is unknown we know just the expected value.

It's wrong to say that (just if you can confirm it will be great):
[itex] E[\frac{1}{X+1}]=\frac{1}{E[X]+1}[/itex]
Thanks and sorry for repost
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mathman
#2
Nov4-11, 03:50 PM
Sci Advisor
P: 6,039
You need the distribution function for X (the mean is not enough) to get the mean of 1/(X+1).
giglamesh
#3
Nov5-11, 03:54 PM
P: 14
thanks apparently I do

bpet
#4
Nov6-11, 05:22 AM
P: 523
Expected Value of reciprocal (Sorry for reposting)

If X is strictly positive, you can apply Jensen's inequality etc. to get 1 >= E[1/(X+1)] >= 1/(E[X]+1).
Stephen Tashi
#5
Nov7-11, 06:17 PM
Sci Advisor
P: 3,252
Quote Quote by giglamesh View Post

There is event Y when calculated it gives the value:

[itex]P(Y)=E[\frac{1}{X+1}][/itex]
What does that notation mean? Is Y some event ( like "A red bird lights on the window") and P(Y) is its probability?
harry311
#6
Apr5-12, 05:48 PM
P: 2
hi giglamesh, have you had the answer so far? Im having exactly problem like you
giglamesh
#7
Apr5-12, 05:56 PM
P: 14
hi all
yes P(Y) is another event which probability is the expected value of other function of random variable.
Applying Jenesen Inequality does not help because it gives the lower bound.
So I decided to work on the problem to get X distribution to calculate the E[1/(1+X)]

but few days later I modified the problem to another distribution P(Y)=E[1/X] in another post.
Greetings
harry311
#8
Apr5-12, 05:59 PM
P: 2
did you get the answer for E(1/X) as well?
giglamesh
#9
Apr5-12, 06:02 PM
P: 14
yes just find the distribution of X, the PMF (discret case)
then calulate the probability like this:
P(Y)=E[1/X]=sum_{i=1}^{i=n}{(1/i)*P(X=i)}
using Jenesen inequality here doesn't help because the funtion is defined to be 0 at 0 so we can't consider it convex.
hope that would help


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