Show that Xbar - Ybar is a consistent estimator

[birdec]

Suppose that X sub 1, X sub 2,... X sub n and Y sub 1, Y sub 2,... Y sub n are independent random samples from populations with means mu sub x and mu sub y and variances sigma squared sub x and sigma squared sub y , respectively. Show that X bar - Y bar is a consistent estimator of mu sub x - mu sub y.

I know that the Bias and Variance must equal 0 so...

Bias (Xbar - Ybar) =
[E(Xbar) - mu sub x] - [E(Ybar) - mu sub y]
= 0 Variance (Xbar - Ybar)
[sigma squared sub x /n] - [sigma squared sub y /n]
= 0

I'm pretty sure this is incorrect.

 

[Stephen Tashi]

I know that the Bias and Variance must equal 0

That isn't what it means for an estimator to be consistent. Why don't you look up the definition?

(If the variance were zero, the random variable would have only one possible value.)