Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 2 of 2
  1. MHB Craftsman

    Status
    Offline
    Join Date
    Mar 2017
    Posts
    151
    Thanks
    42 times
    Thanked
    77 times
    #1
    Let $n$ be a positive integer, and let $G$ be a random graph from $G(n, 1/2)$. Let $e_1, . . . , e_{n \choose 2}$ be the
    possible edges on the vertex set ${1, . . . , n}$, and for each $i$, let $A_i$ be the event that $e_i ∈ E(G)$.
    Prove that the events $A_1, . . . , A_{n \choose 2}$ are independent.


    Can I just have some help understanding what details I should be including here?
    It's so trivial that I don't know how to write it down as a proof.
    There is a probability of $\frac{1}{2}$ that any arbitrary possible edge will be an edge of the graph. And it just seems obvious that whether one edge is in the graph has nothing at all to do with if another edge is in the graph.

    So then if we choose an arbitrary subset of $A_1, . . . , A_{n \choose 2}$, the probability of the events in the subset occurring will be the product of the individual probabilities of the events.

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Journeyman
    MHB Math Scholar
    caffeinemachine's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    India
    Posts
    823
    Thanks
    584 times
    Thanked
    1,149 time
    Thank/Post
    1.396
    Awards
    MHB Topology and Advanced Geometry Award (2016)
    #2
    Quote Originally Posted by joypav View Post
    Let $n$ be a positive integer, and let $G$ be a random graph from $G(n, 1/2)$. Let $e_1, . . . , e_{n \choose 2}$ be the
    possible edges on the vertex set ${1, . . . , n}$, and for each $i$, let $A_i$ be the event that $e_i ∈ E(G)$.
    Prove that the events $A_1, . . . , A_{n \choose 2}$ are independent.


    Can I just have some help understanding what details I should be including here?
    It's so trivial that I don't know how to write it down as a proof.
    There is a probability of $\frac{1}{2}$ that any arbitrary possible edge will be an edge of the graph. And it just seems obvious that whether one edge is in the graph has nothing at all to do with if another edge is in the graph.

    So then if we choose an arbitrary subset of $A_1, . . . , A_{n \choose 2}$, the probability of the events in the subset occurring will be the product of the individual probabilities of the events.
    In $G(n, p)$, the edges appear independently with probability $p$ (by definition of $G(n, p)$). So the question indeed is asking to prove something that is immediate from the definition.

Similar Threads

  1. [SOLVED] Number of edges of the graph
    By mathmari in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 4
    Last Post: August 29th, 2018, 14:49
  2. Securing all edges of an area
    By bio in forum Geometry
    Replies: 6
    Last Post: May 15th, 2018, 12:23
  3. [SOLVED] Choosing a ball at random from a randomly selected box
    By Sapphireluna in forum Basic Probability and Statistics
    Replies: 2
    Last Post: March 22nd, 2016, 10:13
  4. MGF relating to random sum of random variables
    By nedflanders in forum Advanced Probability and Statistics
    Replies: 1
    Last Post: October 15th, 2014, 16:06
  5. Choosing without replacement
    By evinda in forum Basic Probability and Statistics
    Replies: 8
    Last Post: April 14th, 2013, 11:16

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards