Assuming that $H$, $N$ are random variables in which $H$ is distributed following exponential distribution with mean value $\Omega$, and $N$ is a Gaussian random variable with probability density function $$ f_N(x)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right). $$

Let $a$, $b$, and $c$ be constants. We define the following random variables

\begin{align} Y_1 &= |aH+bN|,\\ Y_2 &= |cN|. \end{align} Let $Y=\max\{Y_1, Y_2\}$. Find the CDF of $Y$.

Could you please give me a hint to find the CDF of $Y$. Thank you very much.

  • $\begingroup$ This looks like something I would like to know also. Should you try to find the joint PDF first by trying some kind of transform on Y? I'm half way into my Stochastic modeling class, however I don't seem to remember anything quite like this. $\endgroup$ – Jabernet Mar 21 '15 at 0:53
  • $\begingroup$ You said you were considering the random variables $H,N$ and $M$ but you never said anything about $M$. Is this a typo? Also using $\Omega$ for the mean value of $H$ might not be the best notation. $\endgroup$ – Reveillark Mar 21 '15 at 1:17
  • $\begingroup$ Thank Reveillark for your comment. I have just edited it. $\endgroup$ – Tran Tam Mar 21 '15 at 2:31
  • $\begingroup$ First, the PDF of $N$ is not correct unless $\sigma = 1$ and you should resolve that. Then, because there is not a closed form CDF for normal, it isn't immediately obvious how you can get a closed form CDF for either $Y_i$. $\endgroup$ – BruceET Mar 21 '15 at 19:45
  • $\begingroup$ Thank Prof Bruce Trumbo, I have just edited the PDF of N. Could you please tell me that, it is possible to find the CDF of Y when $\sigma=1$ and please give me a hint to do it. $\endgroup$ – Tran Tam Mar 21 '15 at 23:21

In general, if $Y_1$ and $Y_1$ are random variables and $Y=\max\{Y_1,Y_2\}$, then for any $x\in\mathbb R$, $$\{\omega: Y(\omega)\leqslant x\} = \{\omega :Y_1(\omega)\leqslant x\}\cap\{\omega: Y_2(\omega)\leqslant x\}. $$ So $$\mathbb P(Y\leqslant x) = \mathbb P\left(\{Y_1\leqslant x\}\cap\{Y_2\leqslant x\}\right). $$ If $Y_1$ and $Y_2$ are independent, then this reduces to $$\mathbb P(Y_1\leqslant x)\mathbb P(Y_2\leqslant x), $$ i.e. the product of the distribution functions of $Y_1$ and $Y_2$. If not, then it is a bit more complicated, and you would need to find the joint distribution of $(Y_1, Y_2)$.

  • $\begingroup$ Thank Math1000. But $Y_1$ and $Y_2$ are obvious interdependent. Thus, we can not apply directly your approach. $\endgroup$ – Tran Tam Mar 21 '15 at 23:22
  • $\begingroup$ I posted this more to give some ideas than as a true answer (and it's too long for the comments). I'm not sure how to approach this off the top of my head, sorry. $\endgroup$ – Math1000 Mar 22 '15 at 2:23
  • $\begingroup$ Dear Math1000, thank for your hint. $\endgroup$ – Tran Tam Mar 22 '15 at 4:46

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