I have not idea about part a. I know I need to prove the integration of f(x,y)=1, but how should I deal with the range of x and y.
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We deal with part a), which is causing you some difficulty. It is basically a calculus problem.
Our density function is $0$ except on the part of the $x$-$y$ plane which is below the line $y=x$. Call this region $D$. We want to show that $$\iint_D \frac{2e^{-2x}}{x}\,dx\,dy=1.$$ One standard way to evaluate this kind of double integral is to express it as an iterated integral. We have in principle a choice as to whether to integrate first with respect to $x$ or with respect to $y$. With respect to $x$ looks difficult. So let us integrate first with respect to $y$. We get $$\int_{x=0}^\infty \left(\int_0^x \frac{2e^{-2x}}{x}\,dy\right)\,dx.$$ The inner integral is simply $x\frac{2e^{-2x}}{x}$, which is $2e^{-2x}$. The outer integral is therefore $$\int_{0}^\infty 2e^{-2x}\,dx.$$ This integral is easy to evaluate. It is $1$.
On the way, we have found the (marginal) density of $X$, for that is obtained by "integrating out" $y$. The marginal density is $2e^{-2x}$ (when $x\gt 0$). For $x\le 0$, the density function of $X$ is $0$.
André Nicolas
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You aare welcome. – André Nicolas Oct 22 '14 at 05:09
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for part b, is that we need use the formula of f(x,y)/fx(x) to find ? – kenxilo Oct 22 '14 at 05:23
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Yes, divide the joint density by the marginal density of $X$ calculated in (a), – André Nicolas Oct 22 '14 at 05:30
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http://math.stackexchange.com/questions/985422/a-question-of-joint-cdf can help me have a look of this question... thanks in advance – kenxilo Oct 22 '14 at 06:22
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I looked, that question has been answered. We want $\Pr(X\le x \quad\text{and}\quad Y\le y)$. You get a slightly unpleasant "cases" function. It helps a lot to drae a picture. – André Nicolas Oct 22 '14 at 06:40