# Thread: P(Z<z), Z=X if X<=1, Z=Y+1 if X>1

1. ## P(Z<z), Z=X if X<=1, Z=Y+1 if X>1

So here is the question

X∼exp(1)X
Y∼exp(2)Y (exponentialy distributed)
X and y are independent.

Z=Xif x≤1. Otherwise Z=Y+1. What is the cumulative distribution function of Z?

So what i thought.
Fz(z)=P(Z<z)=P(Z<z∣X<1)×P(X<1)+P(Z<z∣X>1)×P(X>1)=(1−e−z)(1−e−1)+(1−e−2(z−1))

But i am realy not sure. Should z maybe be devided in to tow cases wher z<1 and z>1 maybe? So F(z) is not continouse?

2. ## Re: P(Z<z), Z=X if X<=1, Z=Y+1 if X>1

can anyone tell me if this is possible? please I've been breaking my head over this for 2 days. and I am actualy pretty sure its not suppose to be that hard.

F(z) = (1-e^-z)(1-e^-1) if Z<=1
F(z)=(1-e^-2(z-1))e^-1 if z>1

or
maybe it would be
F(z)=(1-e^-z)(1-e^-1)+(1-e^-2(z-1))e^-1
both of them I derive from the total probability rule but one is divided into 2 parts and on is not.
but both don't feel right and I have no idea how else to approach it.

3. ## Re: P(Z<z), Z=X if X<=1, Z=Y+1 if X>1

The given question is are independent. Assuming the parametrization means the pdf of is . Define

It is trivial that . By law of total probability, for , the CDF

 Tweet

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts