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.

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.

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