Solve Probability Question Homework: Ex, Var X

[GreenPrint]

Homework Statement

Random variable X takes values a,a+Δ,a+2Δ,...,a+nΔ with equal probabilities. (a) What is Ex? (b) What is Var X?

Homework Equations

The Attempt at a Solution

X = [a,a+Δ,a+2Δ,...,a+nΔ]
X = a+(X'-1)Δ
where X' is some random variable
X' = [1,2,...,n+1]
I test to make sure this is true
X = [a+(1-1)Δ,a+(2-1)Δ,a+(3-1)Δ...,a+(n+1-1)Δ]=[a,a+Δ,a+2Δ...,a+nΔ]
So it's true.

E[X] = E[a+(X'-1)Δ] = a+(E[X']-1)Δ

E[X']=$\sum^{n+1}_{X'=1}$X'P(X')

Since X has equal probabilities then X' has equal probabilities so it's just
P(X') = $\frac{1}{number of entities}$
number of entities in this case is n+1 so
E[X']=$\sum^{n+1}_{X'=1}$X'($\frac{1}{n+1})$
So I can pull out the constant from the summation
=$\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}(1+2+...+n+1)$
I know that the quantity 1+2+...+n+1 can be expressed as $1+\frac{(n+1)n}{2}$ so...
$\frac{1}{n+1}(1+\frac{(n+1)n}{2}) = \frac{1}{n+1}+\frac{n}{2}$

however the answer key which my professor made up says that it should be $\frac{n}{2}+1$. However I don't see how my answer is wrong.

Thanks for any help that anyone can provide me in solving this problem. My professor didn't show in the answer key how hey got this and just wrote it without showing any work. I think he may be wrong.

 

[Ray Vickson]

Homework Statement

Random variable X takes values a,a+Δ,a+2Δ,...,a+nΔ with equal probabilities. (a) What is Ex? (b) What is Var X?

Homework Equations

The Attempt at a Solution

X = [a,a+Δ,a+2Δ,...,a+nΔ]
X = a+(X'-1)Δ
where X' is some random variable
X' = [1,2,...,n+1]
I test to make sure this is true
X = [a+(1-1)Δ,a+(2-1)Δ,a+(3-1)Δ...,a+(n+1-1)Δ]=[a,a+Δ,a+2Δ...,a+nΔ]
So it's true.

E[X] = E[a+(X'-1)Δ] = a+(E[X']-1)Δ

E[X']=$\sum^{n+1}_{X'=1}$X'P(X')

Since X has equal probabilities then X' has equal probabilities so it's just
P(X') = $\frac{1}{number of entities}$
number of entities in this case is n+1 so
E[X']=$\sum^{n+1}_{X'=1}$X'($\frac{1}{n+1})$
So I can pull out the constant from the summation
=$\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}(1+2+...+n+1)$
I know that the quantity 1+2+...+n+1 can be expressed as $1+\frac{(n+1)n}{2}$ so...
$\frac{1}{n+1}(1+\frac{(n+1)n}{2}) = \frac{1}{n+1}+\frac{n}{2}$

however the answer key which my professor made up says that it should be $\frac{n}{2}+1$. However I don't see how my answer is wrong.

Thanks for any help that anyone can provide me in solving this problem. My professor didn't show in the answer key how hey got this and just wrote it without showing any work. I think he may be wrong.

$$\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{1}{2}n(n+1).$$

 

[GreenPrint]

But I want to sum to n+1 and not n. How does this effect the answer? I haven't taken calculus 2 in several semesters. I remember some change of base thing.

$\sum^{n+1}_{X'=1}X'=\sum^{n}_{X'=0}(X'-1)$

Is this correct?

 

[GreenPrint]

So now that I think of it using the information you provided

$\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}\sum [1 + 2 + 3 + ... + n+1] = \frac{1}{n+1}(1+[1+ 2+ 3 + ... + n]) = \frac{1}{n+1}(1+\frac{1}{2}n(n+1))=\frac{1}{n+1}+.5 n$

I still get the same answer

 

[Ray Vickson]

So now that I think of it using the information you provided

$\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}\sum [1 + 2 + 3 + ... + n+1] = \frac{1}{n+1}(1+[1+ 2+ 3 + ... + n]) = \frac{1}{n+1}(1+\frac{1}{2}n(n+1))=\frac{1}{n+1}+.5 n$

I still get the same answer

It is still wrong. Start again, and do not bother to look at X'; just go with the original X.

 

[GreenPrint]

What am I doing wrong though? I can't seem to find what it is exactly that I am doing wrong.