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Derivation of sampling distribution Q

Can someone explain this to me? I understand up to where it equates the expectation of the individual independent, normal variables with the same number of means. I don't understand how the expectation operator, applied individually to all these variables - creates the same mean. I understand that they are all from the same same, and have normal distribution - but to me, this indicates that just the probability of each should be the same. The random variables do not necessarily have the same real value, and isnt the expectation operator = the variable * it's probability?
(edited 11 months ago)
Reply 1
Original post by vitc83
Can someone explain this to me? I understand up to where it equates the expectation of the individual independent, normal variables with the same number of means. I don't understand how the expectation operator, applied individually to all these variables - creates the same mean. I understand that they are all from the same same, and have normal distribution - but to me, this indicates that just the probability of each should be the same. The random variables do not necessarily have the same real value, and isnt the expectation operator = the variable * it's probability?

The expected value of a random variable is its mean, becaue its the sum (integral) over all the values of the random variable weighted by the corresponding probability. So in this case as the n random variables X_i are all identically distributed, N(mu, s^2), then
E(X_j) = mu
for all i, pretty much by definition.

Your bold isnt correct, as each X_i is a random variable which represents the values/probability that the sample can be. There is no "real value".
(edited 11 months ago)
Reply 2
Original post by mqb2766
The expected value of a random variable is its mean, becaue its the sum (integral) over all the values of the random variable weighted by the corresponding probability. So in this case as the n random variables X_i are all identically distributed, N(mu, s^2), then
E(X_j) = mu
for all i, pretty much by definition.

Your bold isnt correct, as each X_i is a random variable which represents the values/probability that the sample can be. There is no "real value".


Okay - but surely the values that the sample can be are gonna be different? Like I understand that if they have the same distribution, they'll have the same mean... but like if X_i is a value from a sample, then how do they all have the same mean.
(edited 11 months ago)
Reply 3
Original post by vitc83
Okay - but surely the values that the sample can be are gonna be different? Like I understand that if they have the same distribution, they'll have the same mean... but like if X_i is a value from a sample, then how do they all have the same mean.

I think youre getting confused over what "a sample" means and how the expectation is applied. As a thought experiment, imagine taking a single set of n random values (here theyre drawn from a normal distribution, independent etc), then take a second set of n random values, then a third set of n random variables and so on .... When you calculate E(X_1), this refers to calculating the expected value of the first element in each set, averaged across all the sets. Similarly E(X_2) is the expected value of the second element in each set, averaged over all the sets, .... For X_1, for example, the sample in each set is randomly drawn from N(mu, s^2) so
E(X_1) = mu
and similarly for E(X_2), ....

Its misleading to think of X_1 as having a specific value. Rather its a random variable ~N(mu,s^2) and if you do want to think of it in terms of having specific value(s), then the previous hypothetical case should make sense?
(edited 11 months ago)
Reply 4
j

Original post by mqb2766
I think youre getting confused over what "a sample" means and how the expectation is applied. As a thought experiment, imagine taking a single set of n random values (here theyre drawn from a normal distribution, independent etc), then take a second set of n random values, then a third set of n random variables and so on .... When you calculate E(X_1), this refers to calculating the expected value of the first element in each set, averaged across all the sets. Similarly E(X_2) is the expected value of the second element in each set, averaged over all the sets, .... For X_1, for example, the sample in each set is randomly drawn from N(mu, s^2) so
E(X_1) = mu
and similarly for E(X_2), ....

Its misleading to think of X_1 as having a specific value. Rather its a random variable ~N(mu,s^2) and if you do want to think of it in terms of having specific value(s), then the previous hypothetical case should make sense?


Okay, so I think I've mixed up what X_i; so to clarify, its basically the ith element of the set, which itself is a sample of the population. Multiple samples are taken. So X_1 for an example, is ALL of the first elements of each set?

So the expected value of X_1 is like, the sum of all the first elements multiplied by their probabilities. But how can this give the same mean for every X_1, X_2 etc
Reply 5
Original post by vitc83
j



Okay, so I think I've mixed up what X_i; so to clarify, its basically the ith element of the set, which itself is a sample of the population. Multiple samples are taken. So X_1 for an example, is ALL of the first elements of each set?

So the expected value of X_1 is like, the sum of all the first elements multiplied by their probabilities. But how can this give the same mean for every X_1, X_2 etc

X_i is the ith random variable in the set (or sum). Its a random variable, N(mu, s^2), not a value/"sample".
https://en.wikipedia.org/wiki/Expected_value#Random_variables_with_density
As such, the expected value E(X_i) is simply the mean of the underlying distribution
E(X_i) = mu
and this is the same for all i.

As it says at the start
https://en.wikipedia.org/wiki/Expected_value
"Informally, the expected value is the arithmetic mean of a large number of independently selected outcomes of a random variable"
which is the example I tried to give in the previous post. I assumed you drew vectors/sets of n random values from that set of n random variables and then averaged (take the expected value) over the sets. So for X_1, all the values in the first position in each vector/set are independently drawn from N(mu, s^2) and their average is simply mu which tends to E(X_1). Obviuosly this is the "same" for X_2, X_3, ...

Edit - to try and make it clear, when you start with
X1 + X2 + ... + Xn
what I see it as is roughly
Bellcurve + Bellcurve + ... + Bellcurve
where each of the Bellcurves is N(mu,s^2). Really when you add random variables, the resulting distirbution is found using convolution
https://en.wikipedia.org/wiki/Convolution_of_probability_distributions
However, adding iid normals is simpler and taking expectations you get
E(X1) + E(X2) + ... + E(Xn)
which is
E(Bellcurve) + E(Bellcurve) + ... + E(Bellcurve)
and as all the bellcurves are identical, then by definition of E() this is simply
mu + mu + ... + mu = n*mu
(edited 11 months ago)

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