- #1
Calu
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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!
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!