Re: Distribution functions
Quote:
Originally Posted by
supermario88
Hi there,
I was wondering if someone could explain how to go about the following problem.
Let X and Y be independent random variables having distribution functions Fx and FY , respectively.
a) Define Z = max{X,Y} to be the larger of the two. Show that FZ(z) = FX(z)FY(z) for all z.
b) Define W = min{X,Y} to be the smaller of the two. Show that FW(w) = 1 - [1-FX(w)][1-FY(w)] for all w.
Thanks!
since they are independent,
![F_Z(z)=P[Z \leq z]=P[(X \leq z)\cap (Y \leq z)]= P(X \leq z) P(Y \leq z)=F_X(z)\cdot F_Y(z)](http://latex.codecogs.com/png.latex?F_Z(z)=P[Z \leq z]=P[(X \leq z)\cap (Y \leq z)]= P(X \leq z) P(Y \leq z)=F_X(z)\cdot F_Y(z))
and
![F_W(w)=P[W \leq w]=1-P[(X>w) \cap (Y>w)]=.......](http://latex.codecogs.com/png.latex?F_W(w)=P[W \leq w]=1-P[(X>w) \cap (Y>w)]=.......)
Re: Distribution functions
Hi harish21,
Thanks very much for the reply. So I can finish letter b thanks to your hint. Would you mind explaining the intuition of letter a? Specifically, could you talk about why
P[Z≤z] = P[(X≤z)∩(Y≤z)]
I just want to develop a better understanding of the problem. Thank you.
Re: Distribution functions
Z is the largest of X and Y
so IF Z is less than or equal to a, then both X and Y must be less than or equal to a.
For the minimum, use the complement twice
Re: Distribution functions
Re: Distribution functions
I work with order statistics all the time.
I can google my name and find some of my papers...
http://w3.math.sinica.edu.tw/bulleti...d322/32203.pdf
Re: Distribution functions
Understood. Thanks matheagle.