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Division of Two exponential Random variables

I have two exponential random variables $X$ and $Y$ with rate parameter $\lambda_1$ and $\lambda_2$. I have to find distribution function of $Z$ where $Z$ is given by:

$$Z=\frac{X}{Y+1}$$ and $f_{X}(x)=\lambda_1.\exp(-\lambda_1 x)$ and $f_{Y}(y)=\lambda_2.\exp(-\lambda_2 y)$. How to find pdf of $Z$ and How to check its validity?

I followed this procedure and got an answer as: \begin{eqnarray} F(Z)=p{(Z \leq z)}\\ &=&P\left\{\frac{X}{Y+1} \leq z \right\}\\ &=&P\left\{X \leq z.(y+1) \right\}\\ &=&\int_{y=0}^{+\infty} \int_{x=0}^{z(y+1)}f(x,y) dx\,dy$\\ &=&\int_{y=0}^{+\infty} \int_{x=0}^{z(y+1)}\lambda_1\lambda_2\exp(-\lambda_1x)\exp(-\lambda_2y) dx\,dy\\ &=&\lambda_1\lambda_2\int_{y=0}^{+\infty} \exp(-\lambda_2y)\int_{x=0}^{z(y+1)}\exp(-\lambda_1x)dx\,dy\\ &=&\lambda_2\int_{y=0}^{+\infty}\exp(-\lambda_2y).(1-\exp(-\lambda_1(1+y).z))dy\\ &=&\lambda_2\left(\frac{1}{\lambda_2}-\frac{exp(-\lambda_1z)}{\lambda_2+\lambda_1.z}\right) \end{eqnarray} which is CDF of $F(Z)$. To find pdf $F_{Z}(z)$ I derivated $F(Z)$ and got the result as: \begin{eqnarray} f_{Z}(z)&=&\frac{dF(Z)}{dz}\\ &=&\lambda_1\lambda_2\exp(-\lambda_1z).\left[\frac{1}{\lambda_1 z+\lambda_2}+\frac{1}{(\lambda_1z+\lambda_2)^2}\right] \end{eqnarray}

Now, How do I check it's Validity?

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  • $\begingroup$ Are $X$ and $Y$ independent ? $\endgroup$ – jibounet Sep 15 '17 at 11:29
  • $\begingroup$ yes,$X$ and $Y$ are independent $\endgroup$ – Nagesh Sep 15 '17 at 11:31
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Let $f$ be a continuous real-valued bounded function defined on $[0,+\infty[$. We want to compute $\mathbb{E}\big[ f(Z) \big]$. Given that $X$ and $Y$ are independent (and both follow exponential distributions), we have:

$$ \mathbb{E}\big[ f(Z) \big] = \int_{0}^{+\infty} \int_{0}^{+\infty} f\Big(\frac{x}{y+1}\Big) \lambda_1 e^{-\lambda_{1}x} \lambda_{2}e^{-\lambda_{2}y} \; dxdy. $$

Consider the change of variables $\Phi \, : \, [0,+\infty[^2 \, \rightarrow \, [0,+\infty[^2$ defined by:

$$ \forall (x,y) \in [0,+\infty[^2, \; \Phi\big( (x,y) \big) = \Big(\frac{x}{y+1}, y\Big) = (u,v) \in [0,+\infty[^2. $$

$\Phi$ is a $\mathcal{C}^{1}$ diffeomorphism from $[0,+\infty[^2$ to itself and:

$$ \Phi^{-1}\big( (u,v) \big) = \big( u(v+1), v \big) $$

$$ \mathrm{Jac}\big( \Phi^{-1}, (u,v) \big) = \begin{bmatrix} v+1 & u \\ 0 & 1 \end{bmatrix}. $$

It follows from the change of variable theorem that:

$$ \begin{align*} \mathbb{E}\big[ f(Z) \big] & = {} \int_{0}^{+\infty} \int_{0}^{+\infty} f(u) \lambda_1\lambda_2 e^{-\lambda_{1}u(v+1)}e^{-\lambda_{2}v} (v+1) \, du dv. \\[2mm] & = \int_{0}^{+\infty} f(u) \Bigg( \lambda_1 \lambda_2 e^{-\lambda_1 u} \int_{0}^{+\infty} (v+1) e^{-(\lambda_{1}u + \lambda_{2})v} \, dv \Bigg). \end{align*}$$

Also, note that:

$$ \begin{align*} \int_{0}^{+\infty} (v+1)e^{-(\lambda_{1}u + \lambda_{2})v} dv & = {} \Bigg[-\frac{e^{-av}(av + a + 1)}{a^2} \Bigg]_{v=0}^{v=+\infty} \\ & = \frac{a+1}{a^2}. \end{align*}$$

with $a = \lambda_{1}u + \lambda_{2}$. Therefore, the probability density function of $Z$ (with respect to the Lebesgue measure on $[0,+\infty[$) is given by:

$$ u \in [0,+\infty[ \, \mapsto \; \lambda_{1}\lambda_{2}e^{-\lambda_{1}u} \frac{\lambda_{1}u + \lambda_{2} + 1}{(\lambda_{1}u + \lambda_{2})^2}. $$

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  • $\begingroup$ How to check it's validity? I have got same answer as you. $\endgroup$ – Nagesh Sep 15 '17 at 12:44
  • $\begingroup$ @Nagesh : If there is no mistake in the proof, the answer is valid. If you want, you can check this result numerically: 1) Draw (independently) a very large number of realizations of $X$ and $Y$. Call these realizations $(x_i, y_i)_{1 \leq i \leq N}$. 2) Compute $z_i = x_i / (y_i + 1)$ for all $i$. 3) Plot a normalized histogram of $(z_i)_{1 \leq i \leq N}$ 4) On top of this histogram, plot the density function. $\endgroup$ – jibounet Sep 15 '17 at 12:47
  • $\begingroup$ does property of pdf which is "area under pdf curve is 1" support here? $\endgroup$ – Nagesh Sep 15 '17 at 12:51
  • $\begingroup$ @Nagesh : Yes. The function obtained at the end of my post integrates to 1. Therefore it is actually a PDF. $\endgroup$ – jibounet Sep 15 '17 at 12:54
  • $\begingroup$ But I'm not getting 1 after integrating. What I'm getting is: $1+\lambda_2 (\lambda_2+1) \left(-\log \left(\frac{\lambda_1}{\lambda_2}\right)+\log (\lambda_1)-\log (\lambda_2)\right)$ $\endgroup$ – Nagesh Sep 15 '17 at 12:56

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