If $X_1$ and $X_2$ are independent exponential random variables with respective parameters $\lambda_1$ and $\lambda_2$, find the distribution of $Z=X_1/X_2$. Also compute $P(X_1 \lt X_2)$.
What information does the question want? I know the expectation of $X_1$ is $\lambda_1$ and the expectation of $X_2$ is $\lambda_2$ so the expectation of $Z$ should be $\lambda_1/\lambda_2$ right? and the variance of $X_1$ is $\lambda_1$ and the variance of $X_2$ is $\lambda_2$ so the variance of $Z$ is $\lambda_1/\lambda_2$, is this correct? what other information does the distribution ask for?
and for the $P(X_1 \lt X_2)$ can I solve this with the integral $\int_0^{\infty} \int_0^{X_2} \lambda e^{-\lambda x} dX_1 dX_2$?

  • 2
    $\begingroup$ The expectation of a ratio is not equal to the ratio of the expectations... And in any case, the expectation alone is not enough to determine the distribution. Otherwise, normal random variables with the same mean but different variances would be equal in distribution, which is clearly absurd. $\endgroup$ – M Turgeon Dec 5 '15 at 1:48
  • $\begingroup$ Related $\endgroup$ – A.S. Dec 5 '15 at 2:08
  1. $P(X<Y)$.

$X,Y$ are independent exponentials with rates $\lambda$, $\mu$. Then one way to do it is $$P(X<Y) = \int_0^\infty\int_x^\infty f_{X,Y}(x,y)\,dydx = \int_0^\infty\int_x^\infty f_X(x)f_Y(y)\,dydx$$ because of independence. This is called competing exponentials. The result is $$\frac{\lambda}{\lambda+\mu}.$$

  1. Distribution of $Z = Y/X$.

If $X,Y$ are independent exponentials with rates $\lambda,\mu$, then $Y = ZX$ and one way to do it is \begin{align*} f_Z(z) &=\int_0^\infty f_X(x)f_Y(zx)\left|\frac{dy}{dz}\right|dx\\ &= \int_0^\infty \lambda e^{-\lambda x}\cdot \mu e^{-\mu zx}|x|\,dx\\ &= \int_0^\infty \lambda\mu e^{-(\lambda +\mu z)x}|x|\,dx\\ &= \frac{\lambda\mu}{(\lambda+\mu z)^2}. \end{align*}

Addendum: Using the CDF gives \begin{align*} F_Z(z) &= P(Z<z)\\ &= P(Y/X<z)\\ &= P(Y<zX)\\ &=\int_0^\infty\int_{y/z}^\infty f_X(x)f_Y(y)\,dxdy\\ &=\int_0^\infty\mu e^{-\mu y}\left\{\int_{y/z}^\infty\lambda e^{-\lambda x}\,dx\right\}dy\\ &=\mu \int_0^\infty e^{-(\mu+\lambda/z)y}\,dy\\ &= \frac{\mu}{\mu+\lambda/z}. \end{align*} I specifically chose those bounds to cut down on steps, recognizing the use of the survival of an exponential. Taking the derivative gives $f_Z(z)$.

  • $\begingroup$ where did the absoule value x come from in the distribution of Z? and why did you add a z in the exponent for $\mu$ $e^{-\mu zx}$? $\endgroup$ – idknuttin Dec 5 '15 at 2:45
  • 1
    $\begingroup$ You might be more familiar with the alternate method I posted. $\endgroup$ – Em. Dec 5 '15 at 3:32

There is no reason why the expectation of $X_1/X_2$ should be $\lambda_1/\lambda_2$. Expectations don't work like that. Similarly for the variance.

When they ask you to find the distribution of $Z$, they want you to determine $P(Z\leq x)$ for an arbitrary real number $x$. This is the distribution function of $Z$.

As for $P(X_1<X_2)$, you need to integrate the joint density of $(X_1,X_2)$, and it looks like you only have the density for one of them. Since they are independent, the joint density is the product of the densities.

  • $\begingroup$ for $P(X_1 \lt X_2)$ since they are both exponential functions should the integral be $\int_0^{\infty} \int_0^{X_2} \lambda_1 e^{-\lambda_1 x} \lambda_2 e^{-\lambda_2 x}dX_1 dX_2$? $\endgroup$ – idknuttin Dec 5 '15 at 2:05
  • 1
    $\begingroup$ Apart from cap $X$ where lower case is appropriate, you have a right integral. The version by peobablyme is a little easier to evaluate. $\endgroup$ – André Nicolas Dec 5 '15 at 2:54

We have \begin{align} \mathbb P(X<Y) &= \mathbb P(\omega\in\Omega: X(\omega)<Y(\omega)\\ &= \mathbb P(\omega\in\Omega : (X,Y)\in\{(x,y)\in(0,\infty)^2 : x<y\})\\ &= \mathbb P\circ(X,Y)^{-1}(\{ (x,y)\in(0,\infty)^2 : x<y\})\\ &= \iint\limits_{x<y} f_{X,Y}(x,y)\ \mathsf d(x\times y). \end{align} Since $f_{X,Y}>0$ a.s. and $$\iint f_{X,Y}\ \mathsf d(x\times y)=1<\infty $$ by Tonelli's theorem we may evaluate the double integral as an iterated integral: \begin{align} \iint\limits_{x<y} f_{X,Y}(x,y)\ \mathsf d(x\times y) &= \int_0^\infty \int_x^\infty f_{X,Y}(x,y)\ \mathsf d y\ \mathsf dx. \end{align} By independence, $$f_{X,Y}(x,y) = f_X(x)f_Y(y) = \lambda e^{-\lambda x}\mu e^{-\mu y} $$ and so the probability is $$\int_0^\infty\int_x^\infty \lambda e^{-\lambda x}\mu e^{-\mu y}\mathsf dy\ \mathsf dx = \frac\lambda{\lambda+\mu}. $$ If $Z=\frac YX$ then for $z>0$, $$Z\leqslant z\iff \frac YX\leqslant z\iff Y\leqslant zX. $$ Note that $zX\sim\operatorname{Exp}\left(\frac\lambda z\right)$, as $$\mathbb P(zX\leqslant x) = \mathbb P\left(X\leqslant \frac xz\right) = 1-e^{-\frac\lambda z x}, $$ so $$\mathbb P(Z\leqslant z) = \mathbb P(Y\leqslant zX) = \frac\mu{\mu+\frac\lambda z}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.