density functionintegrationprobability
Find the density function of $Z=XY$.
I was given $f(x,y)$ for $x>0$, $y>0$.
I know $F_{z}(z)= \iint f(x,y)dxdy$.
But I need help finding the bounds for the double integral.
I think the first one goes from $0$ to infinity and the second one goes from $0$ to z, but I am unsure.
Thank you!
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 $\{(g,m)\in\Bbb R^2: g\geq 2m, m>0\}$ $$\mathsf P(G\geqslant 2M)=\int_0^\infty \int_{2m}^\infty 0.18\mathsf e^{-0.3g-0.6m}~\mathsf d g~\mathsf d m$$
The question is ambiguous and pdf is valid if you choose different $k$ for $y \leq 1$ and $y \geq 1$.
Probability density function $f(x, y) = k (1-y), 0 \leq x \leq y \leq 2$
i) For $0 \leq y \leq 1$
$\displaystyle \int_0^1 \int_0^y k (1-y) \ dx \ dy = \frac{k}{6}$
ii) For $1 \leq y \leq 2$
$\displaystyle \int_1^2 \int_0^y k(1-y) \ dx \ dy = -\frac{5k}{6}$
That leads to two different pdf's for $0 \leq y \leq 1$ (with $k = 1$) and for $1 \leq y \leq 2$ (with $k = -1$). It could have been simply defined as $|1-y|$.
To find probability for the shaded region, split the integral as,
$\displaystyle \int_{0.5}^{0.75} \int_0^y (1-y) \ dx \ dy + \int_{0.75}^{1} \int_0^{0.75} (1-y) \ dx \ dy + \int_{1}^{2} \int_0^{0.75} (y-1) \ dx \ dy$
Best Answer
We have
$$P(XY<Z)=P(X<\frac{Z}{Y})=F(\frac{Z}{Y})$$
Now you simply differentiate
$$\frac{d}{dz}F=\int_0^{\infty}\frac{d}{dz}\int_0^{\frac{Z}{Y}}f(x,y)dxdy$$
By Leibnitz integral rule we have $$\frac{d}{dZ}F=\int_0^{\infty}\frac{f(\frac{Z}{y},y)}{y}dy$$
Where $$\frac{d}{dZ}F=p_z(z)$$
Note that the absolute value in the term $\frac{1}{y}$ is not required, as our domain is $x>0,y>0$.