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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(7.3

^{k}/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(7.3

^{k+1}/(k+1)!) / r(7.3

^{k}/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}P

_{k}= 1 => r∑

^{∞}

_{k=o}(7.3

^{k}/k!) = e

^{7.3}= 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}(7.3

^{k}/k!) = e

^{-7.3}(7.3)∑

^{∞}

_{k=1}7.3

^{k-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}7.3

^{j}/(j)! = e-

^{7.3}(7.3)e

^{7.3}= 7.3

--> This last line I understand also.

Any help you can offer would be greatly appreciated!