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If X1/X2are independent exponential random variables with respective parameters λ1and λ2, find the distribution of Z=X1/X2. Also compute PX1<X2.

Short Answer

Expert verified

The distribution of Z is fz(z)=λ1λ2(λ1z+λ2)2

P(X1<X2)=λ1λ1+λ2

Step by step solution

01

Given information

X1 and X2 are independent exponential random variables.

Parameters λ1andλ2

The joint density function is

fx1x2x,y=λ1e-4xλ2e-λ2x for x, y>0

02

Explanation

The joint density function:

FZz=P(Z<z)=PX1X2<z=PX1<zX2=∫0∞∫02yλ1e-λxλ2e-iydxdy=∫0∞(1-e-4z)λ2e-λydy=1-λ2λ1z+λ2=λ1z+λ2-λ2λ1z+λ2=λ1zλ1z+λ2

The distribution of Z is

f2z=ddzFz(z)=ddzλ1zλ1z+λ2=λ1λ2(λ1z+λ2)2

hence the distribution of z is

fz(z)=λ1λ2(λ1z+λ2)2

Compute P (X1 < X2)

P(X1<X2)=PX1X2<1=P(Z<1)=λ1λ1+λ2

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