Distribution of two independent exponential random variables

  • Thread starter lizzyb
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Q: If [tex]X_1[/tex] and [tex]X_2[/tex] are independent exponential random variables with respective parameters [tex]\lambda_1[/tex] and [tex]\lambda_2[/tex], find the distribution of [tex]Z = X_1 / X_2[/tex].

Discussion

The best method to attack this problem apparent to me is coming up with a cumulative distributive function for [tex]Z[/tex] and then differentiating it.

Work Completed So Far
(note: I don't mean for anyone to check if I did my math right I'm just wondering if I'm going about solving the question right)

[tex]F_Z (a) = P\{ Z \leq a \} = P\{ X_1 / X_2 \leq a \} = P \{ X_1 \leq a X_2 \}[/tex]
[tex]\int_0^\infty \int_0^{a x_2} \lambda_1 e^{- \lambda_1 x_1} dx_1 \lambda_2 e^{- \lambda_2 x_2} dx_2 = \lambda_1 \lambda_2 \int_0^\infty e^{\lambda_2 x_2} dx_2 \left[ -\frac{1}{\lambda_1} e^{-\lambda_1 x_1} \right]^{a x_2}_0 = - \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 ( e^{-\lambda_1 a x_2} - 1 )[/tex]

[tex] = \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 -\lambda_2 \int_0^\infty e^{-\lambda_2 x_2 - \lambda_1 a x_2} dx_2 = \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 - \lambda_2 \int_0^\infty e^{-x_2(\lambda_2 + \lambda_1 a)} dx_2[/tex]

[tex]= \lambda_2 ( - \frac{1}{\lambda_2} ) \left[ e^{- \lambda_2 x_2 } \right]_0^\infty - \lambda_2 \frac{-1}{\lambda_2 + \lambda_1 a} \left[ e^{-x_2(\lambda_2 + \lambda_1 a) \right]_0^{\infty} = 1 - \frac{\lambda_2}{\lambda_2 + \lambda_1 a} [/tex]

So [tex]f_{X_1/X_2}(a) = f_Z(a) = \frac{d F_Z(a)}{da} = \frac{\lambda_1 \lambda_2}{(\lambda_1 a + \lambda_2)^2}[/tex]

Does this look like I'm doing it right?

Thanks.
 
Last edited:

Answers and Replies

  • #2
I'm not an expert in this, but your reasoning all makes sense to me and your math does check out.
 

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