Solve Probability Question: Find r & E[X] = 7.3

In summary: You substituted ##k+1## for ##j## and then took out the constant ##e^{-7.3}(7.3)##. I get it now. Thanks for your help!In summary, the conversation discusses a question about finding the most likely value of X based on a given probability distribution. The answer involves finding the ratio P(X = k+1)/P(X = k) and using it to determine the most likely value of k, which is found to be k = 7. The value of r is then found and used to prove that the expected value of X is 7.3. The conversation also includes a discussion about the summation notation and how constants can be removed from the sum.
  • #1
Calu
73
0
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!
 
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  • #2
Calu said:
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.

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)##?

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.

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.

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!
 
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  • #3
Ahh, I now understand, thanks!
 

1. What is the value of r in this probability question?

Without further context, it is impossible to determine the value of r. More information or equations are needed to solve for r.

2. How can I solve for r in this probability question?

To solve for r, you need to have the necessary equations or formulas related to the probability question. You can then use algebraic manipulation or statistical methods to solve for r.

3. What does E[X] = 7.3 mean in this probability question?

E[X], or expected value, is a measure of the average outcome of a random variable. In this case, it means that the average value of the random variable X is 7.3.

4. How is E[X] related to r in this probability question?

Without further context, it is difficult to determine the relationship between E[X] and r. However, in some cases, r may be used as a parameter in the calculation of E[X].

5. Is there a specific method to solve this probability question?

There are various methods and techniques used to solve probability questions, such as using combinatorics, probability distributions, and statistical methods. The specific method to use depends on the given information and the type of probability question being asked.

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