What Is the Relationship Between the Expected Value of Y and X?

Click For Summary
SUMMARY

The discussion centers on the relationship between the expected value of a random variable X and the expected value of a function of X, specifically Y = E[1/(X+1)]. It is established that knowing only the expected value \bar{X} is insufficient to determine \bar{Y} without the probability mass function (PMF) of X. Participants confirm that Jensen's inequality provides a lower bound for E[1/(X+1)], indicating that E[1/(X+1)] is greater than or equal to 1/(E[X]+1) when X is strictly positive. The conversation also highlights the necessity of determining the distribution of X to compute E[1/X].

PREREQUISITES
  • Understanding of random variables and expected values
  • Familiarity with probability mass functions (PMF)
  • Knowledge of Jensen's inequality and its applications
  • Basic concepts of probability theory
NEXT STEPS
  • Research the derivation and applications of Jensen's inequality in probability
  • Learn how to derive the PMF of a random variable from its expected value
  • Explore methods for calculating expected values of functions of random variables
  • Study the implications of convex functions in probability distributions
USEFUL FOR

Mathematicians, statisticians, data scientists, and anyone involved in probability theory or statistical analysis who seeks to understand the relationships between expected values and random variables.

giglamesh
Messages
14
Reaction score
0
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 \bar{X}

There is event Y when calculated it gives the value:

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

The QUESTION: Is there a way to find expected value \bar{Y} in the terms of \bar{X}? 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):
E[\frac{1}{X+1}]=\frac{1}{E[X]+1}
Thanks and sorry for repost
 
Last edited by a moderator:
Physics news on Phys.org
You need the distribution function for X (the mean is not enough) to get the mean of 1/(X+1).
 
thanks apparently I do
 
If X is strictly positive, you can apply Jensen's inequality etc. to get 1 >= E[1/(X+1)] >= 1/(E[X]+1).
 
giglamesh said:
There is event Y when calculated it gives the value:

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

What does that notation mean? Is Y some event ( like "A red bird lights on the window") and P(Y) is its probability?
 
hi giglamesh, have you had the answer so far? I am having exactly problem like you
 
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
 
did you get the answer for E(1/X) as well?
 
Last edited by a moderator:
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
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K