# Math Help - Shifted exponential

1. ## Shifted exponential

Let X1, ..., Xn be iid r.v. with common pdf
f(x) = e^-(x-θ) , x>θ , -inf < θ < inf
= 0 elsewhere

This pdf is called the shifted exponential. Let Yn = min{X1,...,Xn}. Prove that Yn --> θ in probability, by obtaining the cdf and the pdf of Yn.

2. I not only gave this lecture today I work with order stats.
so far I have ...

Here it is....
Forget about pdf's, you can differentiate the cdf if you wish.
F(x)=1-exp{-(x-theta)} if x exceeds theta and zero otherwise.

So the cdf of Y_n is (I'm using < since it's continuous and I don't want to write <=)
G(y)=P(Y_n<y) = 1-P(Y_n>y)=1-[1-F(y)]^n
=1- exp{-n(x-theta)}
THUS
P(|Y_n-theta|>espislon)=P(theta-epsilon<Y_n<theta+epsilon)
=P(theta<Y_n<theta+epsilon) due to the support
=G(theta+epsilon)-G(theta)
=(1- exp{-n(theta+epsilon-theta)})-(1- exp{-n(theta-theta)})
=1-exp{-n epsilon} -1 +1
=1-exp{-n epsilon}
->1
for all epsilon as n-> infinity