Help with proof that expected value of x_i is the populaton mean X bar

S

sid9221

New Member
#1
#1
I'm having a little trouble with the proof that the expected value of \( x_i \) is \( \bar{X} \).

What I have is

\( E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j) \)

Then

\( Pr(x_i=X_j) = 1/N \)

This is the bit I can't understand, how does that probability evaluate to that value.

I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.

Any advice ?
 
BGM

BGM

TS Contributor
#2
#2
Sorry not understand your question.

Usually in probability notation, upper case letter \( X \) is reserved to random variable while lower case letter \( x \) is for constant.

Please try to state the assumptions and model behind the question. What is given and what is asking for.
 
S

sid9221

New Member
#3
#3
\( \bar{x}\) is the sample mean and \( \bar{X} \) is the population mean.

I'm trying to work through the proof that \( E[\bar{x}]=\bar{X} \)

Hence, I need to evaluate \( E[x_i] \)

Which is where I get the expression I can't understand above.
 
BGM

BGM

TS Contributor
#4
#4
Do you mean

\( X_1, X_2, \ldots, X_n \) are random variables with a common population mean \( \mu \)

You are asked to show the expected value of sample mean \( \bar{X} \), which is defined to be

\( \bar{X} \triangleq \frac {1} {n} \sum_{i=1}^n X_i \)

is equal to the population mean \( \mu \)?

i.e.

\( E[\bar{X}] = \mu \)


BTW what is the source of that question? The notation / wordings are asked quite badly.
 
A

ANDS!

New Member
#7
#7
In these situations it's always best to write out exactly what \(x_{i}\) actually is. In this case \(x_{i} =\frac{ \sum_{j=1}^{n} X_{ij}}{n}\) (assuming the i is meant to indicate which cluster we're in). From there is simply a matter of distributing the expectation over the summation.
 
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