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

ogry

New Member
#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?
I am lost. Please help.
 
Last edited:

ogry

New Member
#2
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.
 
Last edited:

BGM

TS Contributor
#3
The given question is [math] X \sim \text{Exp}(1), Y \sim \text{Exp}(2), X, Y [/math] are independent. Assuming the parametrization means the pdf of [math] Y [/math] is [math] f_Y(y) = 2e^{-2y}, y > 0 [/math]. Define

[math] Z = \begin{cases} X &\text{if } X \leq 1 \\
Y + 1 & \text{if } X > 1 \end{cases} [/math]

It is trivial that [math] \Pr\{Z \leq 0\} = 0 [/math]. By law of total probability, for [math] z > 0 [/math], the CDF [math] F_Z(z) [/math]

[math] = \Pr\{Z \leq z\} [/math]

[math] = \Pr\{Z \leq z, X \leq 1\} + \Pr\{Z \leq z, X > 1\} [/math]

[math] = \Pr\{X \leq z, X \leq 1\} + \Pr\{Y + 1 \leq z, X > 1\} [/math]

[math] = \Pr\{X \leq \min\{z, 1\}\} + \Pr\{Y \leq \max\{z - 1, 0\}\}\Pr\{X > 1\} [/math]

[math] = 1 - e^{-\min\{z, 1\}}
+ \left(1 - e^{-2(z - 1)}\right)e^{-1}\mathbf{1}\{z > 1\}[/math]

[math] = \begin{cases} 1 - e^{-z} & \text{if } 0 < z \leq 1 \\
1 - e^{-1} + e^{-1} - e^{-2z + 1} = 1 - e^{-2z + 1} & \text{if } z > 1
\end{cases} [/math]