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In this question we have that $G \sim \operatorname{Exp}(0.3)$ and $M \sim \operatorname{Exp}(0.6)$ where $G$ and $M$ are independent of each other. I need to find the joint density function and then find $P(G>2M)$.

So far I have that the joint density function is

\begin{align} & f_{GM}(g,m)=f_G(g)f_M(m) \\[10pt] = {} & (0.3e^{-0.3g})(0.6e^{-0.6m}) \\[10pt] = {} & 0.18e^{-0.3g-0.6m} \text{(and $0$ otherwise??)} \end{align}

If this is correct then what is the domain of the density function? If it is $h,m>0$ then I'm not sure what the limits for each of my integrals are. I believe to find $P(G>2M)$ I need to solve a double integral like this:

$$\iint 0.18e^{-0.3g-0.6m} \, dg \, dm$$

...but I don't know how to find the limits of integration. Thanks for any input.

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  • $\begingroup$ Why do you use "displaystyle" instead of an actual display, when nothing but an integral is on one line? $\endgroup$ Commented Sep 17, 2018 at 23:07
  • $\begingroup$ @MichaelHardyHaha I don't know, I'm not very experienced with MathJax so I just utilise the little bits and pieces I know until it looks decent, but not really sure what I'm doing at the same time $\endgroup$
    – Sonjov
    Commented Sep 17, 2018 at 23:26

2 Answers 2

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It is the cartesian product of the support domains of the marginals. $\{(g,m)\in\Bbb R^2: g\geq 0, m\geq 0\}$

$$f_{G,M}(g,m) =0.18 \mathsf e^{-0.3g-0.6m}~\mathbf 1_{g\geq 0, m\geq 0}$$

And of course the domain of the integration for the event of $G\geqslant 2M$ will be $\{(g,m)\in\Bbb R^2: g\geq 2m, m>0\}$ $$\mathsf P(G\geqslant 2M)=\int_0^\infty \int_{2m}^\infty 0.18\mathsf e^{-0.3g-0.6m}~\mathsf d g~\mathsf d m$$

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  • $\begingroup$ Perfect, thanks so much. $\endgroup$
    – Sonjov
    Commented Sep 17, 2018 at 23:58
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The inequalities $g>2m$ and $m < \dfrac g 2$ both say the same thing. So we have: \begin{align} & \iint\limits_{(g,m)\,:\,g\,>\,2m} \cdots\cdots\cdots \, dg \, dm \\[6pt] = {} & \int_0^\infty\left( \int_{2m}^\infty \cdots\cdots\cdots \, dg \right) \,dm \\[25pt] & \iint\limits_{(g,m)\,:\,m\,<\,g/2} \cdots\cdots\cdots \, dg \, dm \\[6pt] = {} & \int_0^\infty \left( \int_0^{g/2} \cdots\cdots\cdots \, dm \right) \, dg \end{align}

It may be simpler one way than the other, but both are correct.

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  • $\begingroup$ Thanks Michael! $\endgroup$
    – Sonjov
    Commented Sep 17, 2018 at 23:59

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