1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability Question

  1. May 17, 2014 #1
    Could someone talk me through this question please? I understand some of it, but I'd like some help understanding the rest. Comments are below each line of the answer.

    Here's the question:

    Evaluate P(X = k+1)/P(X = k) and hence find the most likely value of X when

    P(X = k) = r(7.3k/k!) , for all k = 0, 1, 2 ... ; r > 0.

    Find the value of r and prove that E[X] = 7.3 (where E[X] is the expected value of X).

    Answer:

    P(X = k+1)/P(X = k) = r(7.3k+1/(k+1)!) / r(7.3k/k!) = 7.3/(k+1).

    --> This line I understand just fine.

    ∴ P(X = k+1) > P(X = k) ⇔ 7.3/(k+1) > 1 ⇔ k < 6.3

    --> What I don't understand here is why we can state that P(X = k+1) > P(X = k) ⇔ 7.3/(k+1).

    Hence increasing for k = 0, 1, 2, ... , 6 and P(7) > P(6) and decreasing for k = 7, 8, 9, ...
    ∴ Most likely value for k is k = 7.

    --> I understand the increasing and decreasing parts, but not the P(7) > P(6) or the most likely value of k.

    k=o Pk = 1 => r∑k=o (7.3k/k!) = e7.3 = 1 r = e-7.3.

    --> This line I understand fine.

    From here, I'm struggling to see what's happening:

    E[X] = ∑k=o ke-7.3(7.3k/k!) = e-7.3(7.3)∑k=1 7.3k-1/(k-1)!

    --> I believe by finding the expected value we are finding the total sum of the weighted means, hence the summation has the form ∑k=o k.f(k). However I don't see how the k multiplying the function of k, f(k), has been removed and how the k=1 appears on the summation and k-1 appears in the function, and how the 7.3 finds itself outside the summation as a constant.

    Let j = k+1,

    e-7.3(7.3)∑j=0 7.3j/(j)! = e-7.3(7.3)e7.3 = 7.3

    --> This last line I understand also.

    Any help you can offer would be greatly appreciated!
     
  2. jcsd
  3. May 17, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Using ##p(k)=P(X=k)## you have shown ##p(k+1)>p(k)## if ##k<6.3##. In particular, with ##k=6## this says ##p(7) > p(6)##. And you know ##p(k+1) < p(k)## if ##k\ge 7##. So wouldn't that make ##p(7)## the largest value of any ##p(k)##?

    You have $$\sum_{k=0}^\infty ke^{-7.3}\frac{7.3^k}{k!}$$Notice that the first term when ##{k=0}## is zero so you have$$
    E(X) = \sum_{k=1}^\infty ke^{-7.3}\frac{7.3^k}{k!}$$Also, since ##\frac k {k!} = \frac 1 {(k-1)!}##we have$$
    E(X) = \sum_{k=1}^\infty e^{-7.3}\frac{7.3^k}{(k-1)!}=
    e^{-7.3}(7.3)\sum_{k=1}^\infty \frac{7.3^{k-1}}{(k-1)!}$$The two constants can be removed from the sum and it is now ready for your next step.

     
  4. May 17, 2014 #3
    Ahh, I now understand, thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Probability Question
  1. A probability question (Replies: 3)

  2. Probability Question (Replies: 12)

Loading...