# Minimum of two random variable

Hello, want to know if it's correct
1. Homework Statement

X and Y two random variables iid of common density f and f(x)=x*exp(-x²/2) if x≥0 and f(x)=0 if x≤0
and Z=min(X,Y)
Find
-The density of Z
-The density of Z²
- E[Z²]

## The Attempt at a Solution

1.[/B]
FZ(u) = P(min(X,Y )≤u)
= 1 − P(min(X,Y )>u)
= 1 − P(X>u)*P(Y>u)
= 1 − (1 − F(u))²

And F(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
So FZ(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
The density is then
fZ(u) = u*exp(-u²/2) 0 if u>0 , 0 if u≤0

2. I'm not sure here
u≤0
FZ²(u)=P(Z²≤u)=0
u≥0
Z²≤u , -√u≤Z≤√u
From -√u to 0 , FZ²(u)=0
From 0 to √u , FZ²(u)=FZ(u)=1-exp(-u²/2)

So Z² has the same density as Z ?

Thanks

RUber
Homework Helper

## The Attempt at a Solution

1.[/B]
FZ(u) = P(min(X,Y )≤u)
= 1 − P(min(X,Y )>u)
= 1 − P(X>u)*P(Y>u)
= 1 − (1 − F(u))²

And F(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
So FZ(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
The density is then
fZ(u) = u*exp(-u²/2) 0 if u>0 , 0 if u≤0
I am not following how you went from F(u) to FZ(u) since FZ(u) = 1-(1-F(u))^2. This would give you 1-exp(-u²).
Then fZ(u) = d/du (FZ(u)) = 2u exp(-u^2).
From 0 to √u , FZ²(u)=FZ(u)=1-exp(-u²/2)
You are looking for the probability that Z < sqrt(u), so FZ^2*=(u) = FZ(sqrt(u)) .

Ray Vickson
Homework Helper
Dearly Missed
Hello, want to know if it's correct

FZ(u) = P(min(X,Y )≤u)
= 1 − P(min(X,Y )>u)
= 1 − P(X>u)*P(Y>u)
= 1 − (1 − F(u))²

And F(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
So FZ(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
The density is then
fZ(u) = u*exp(-u²/2) 0 if u>0 , 0 if u≤0

2. I'm not sure here
u≤0
FZ²(u)=P(Z²≤u)=0
u≥0
Z²≤u , -√u≤Z≤√u
From -√u to 0 , FZ²(u)=0
From 0 to √u , FZ²(u)=FZ(u)=1-exp(-u²/2)

So Z² has the same density as Z ?

Thanks

(i) Standard textbook expression: ##f_Z(z) = 2 f(z) P(X>z)##, and in your case, ##P(X > z) = e^{-z^2/2}##. Your final expression is incorrect.
(ii) Since ##Z \geq 0##, ##P(Z^2 \leq w) = P(Z \leq \sqrt{w})## for all ##w \geq 0##. Get the density of ##W = Z^2## by differentiation of ##P(W \leq w)##.