# Probability Question

1. May 17, 2014

### Calu

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.

2. May 17, 2014

### LCKurtz

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.

3. May 17, 2014

### Calu

Ahh, I now understand, thanks!