What is the most general distribution for which E[1/x] = 1/E[x]?

Noe Cowan 2022-11-02 Answered
What is the most general distribution for which E [ 1 / x ] = 1 / E [ x ]?
What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?
Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where E [ 1 / k ] = 1 / E [ k ]. ( k i is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.
So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where E [ 1 / k n ] = 1 / E [ k n ] may also help unifying the models.
What I do know so far, that k i = 1 is a particular solution. In the continuous case every function where f ( x ) = f ( 1 / x ) and E [ x ] = 1 is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.
What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.
You can still ask an expert for help

Want to know more about Discrete math?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Kalmukujobvg
Answered 2022-11-03 Author has 14 answers
Explanation:
By Jensen's inequality applied to the convex function f ( x ) = x 1 on ( 0 , ),
1 E [ X ] < E [ 1 X ]
for any non-constant nonnegative random variable X. Thus, the constant random variable is the only such random variable.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2020-11-09
Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.
asked 2021-08-15
How many elements are in the set { 0, { { 0 } }?
asked 2021-08-18

Discrete Mathematics Basics

1) Find out if the relation R is transitive, symmetric, antisymmetric, or reflexive on the set of all web pages.where (a,b)R if and only if 
I)Web page a has been accessed by everyone who has also accessed Web page b.
II) Both Web page a and Web page b lack any shared links.
III) Web pages a and b both have at least one shared link.

asked 2021-07-28

Let A, B, and C be sets. Show that (AB)C=(AC)(BC)
image

asked 2022-11-17
Counting words of length n from k-sized alphabet with no substring of k consecutive distinct letters
How many words of length n are there, if we have an alphabet of k distinct letters, but the words cannot contain any substring that is made of k consecutive distinct letters, i.e, no k-length substring that consists of the entire alphabet?
Other than that, no restrictions apply: any amount of distinct letters may be used throughout the entire word, and any letter can be used as many times as we like, as long as every substring inside the length n word complies with the rules above.
There is the obvious case of k = 2 which results in 2 words, for every n, because you can only start with either letter, and they alternate.
For the larger case, I have come up with a recursive formula:
C ( n , k ) = f ( 0 , 0 , k )
f ( i , d , k ) = { ( k d ) f ( i + 1 , d + 1 , k ) + c = 1 d f ( i + 1 , c , k ) i < n , d < k 1 i = n , d < k 0 i > n 0 d k
With d being the current length of consecutive distinct letters, and i the current word length.
At every step, we can either use a letter other than the previous d letters, in which case the length is increased by one, and the chain-length d of distinct consecutive letters is also increased by one. In this case, there are ( k d ) such letters we can use at the stage, each subsequently resulting in the same contribution.
Or, we can use a letter already in the last d letters. In this case, the position of the letter matters. If we use the last of the d letters, a whole new chain begins. If instead we use the second last d letter, then a chain of length 2 of consecutive distinct letters begins. The 3rd last would result in a chain of length 3 and so on.
I was wondering if there is a another way to count the amount of such words, perhaps using matrices or combinatorics?
asked 2022-07-08
Inverse Discrete Fourier Transform equation problem
How can I obtain de inverse discrete Fourier Transform of the following equation: X ( j ω ) = k = ( 1 ) k δ ( ω k π 2 )
I have tried by expanding its Fourier series and then doing the inverse and by tables, but I end up obtaining different results, and I am unsure which is the correct one. A step-by-step solution would be very helpful.
Solution 1 (by Fourier Series)
The signal could be written by its Fourier series in this way X ( j ω ) = 1 π k = ( 1 e j k π ) e j k 2 ω
Doing the inverse by definition x ( n ) = 1 2 π π π X ( j ω ) e j ω n d ω
Obtaining x ( n ) = 1 π k = ( δ ( n + 2 k ) e j k π δ ( n + 2 k ) )
Solution 2 (by table)
x ( n ) = 1 2 π k = 0 N 1 ( 1 ) k e j k n π 2
asked 2022-09-05
Discrete Math - non-negative intergers
Find the number of non-negative integer solutions to
x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 3

New questions

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question