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Probability distribution question?
Let X1, X2, ..., Xn be independent normal random variables such that X_i has a normal distribution with mean mu_i and variance sigma_i squared.
a.) Find the distribution of
Y = (sum from i=1 to n) a_i X_i where a1, a2, ..., a_n are fixed constants.
b.) If X1, X2, ..., Xn are independent normal r.v.'s with the same mean mu and variance sigma squared, find the distribution of
X bar = [((sum from i=1 to n) X_i) / (n)]
Sorry about the ugly formatting here, but could someone point me in the right direction here? I think the solution might involve moment-generating functions? The question doesn't really seem too difficult and I can more or less understand what it's asking conceptually, I'm just not sure how to go about solving it.
1 Answer
- IanLv 77 years agoFavorite Answer
Note the following four properties:
1) Any linear combination of independent normally distributed random variables is also normally distributed.
2) The mean of a linear combination of any random variables (independent or dependent) is the linear combination (using the same coefficients) of their means.
3) If a random variable is multiplied by a constant, then its variance is multiplied by the square of that constant.
4) If independent random variables are multiplied by constants, then the variables remain independent.
5) The sum of the variances of independent random variables is the variance of their sum.
Now we are ready to solve the problems.
a) Y = (sum from i=1 to n) a_i X_i is normally distributed with mean
E((sum from i=1 to n) a_i X_i) = (sum from i=1 to n) a_i E(X_i)
= (sum from i=1 to n) a_i mu_i
and variance
Var((sum from i=1 to n) a_i X_i) = (sum from i=1 to n) Var(a_i X_i)
= (sum from i=1 to n) (a_i)^2 Var(X_i)
= (sum from i=1 to n) (a_i)^2 (sigma_i)^2.
b) This is a special case of a), using a_i = 1/n, mu_i = mu, and (sigma_i)^2 = sigma^2 for all i. Therefore, from the result of part a),
X bar = [((sum from i=1 to n) X_i) / (n)] is normally distributed with mean
(sum from i=1 to n) (1/n)mu = n(1/n)mu = mu
and variance
(sum from i=1 to n) (1/n)^2 sigma^2 = n(1/n)^2 sigma^2 = (sigma^2)/n.
By the way, this proof is the reason why, in statistics, the sample mean of n observations has mean mu and standard deviation sigma/sqrt(n).
Have a blessed and wonderful day!
