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Homework Help: 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).


    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


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    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!
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