here's the problem I'm currently working on :
Let $X$ and $Y$ be two continuous random variable with join density function given by $$ f_{X,Y}(x,y)=e^{-x^2y}1_{[1,\infty)}(x)1_{[0,\infty)}(y). $$
Since $e^{x}>0, \forall x \in \mathbb{R}$, we have that $f_{X,Y}>0, \forall (x,y) \in \mathbb{R}^2$. Also we see that
\begin{align*} \int \int_{\mathbb{R}^2} f_{X,Y}(x,y)dxdy &=\int_{1}^{\infty} \left( \int_{0}^{\infty} e^{-x^2y} dy \right) dx\\ &=\int_{1}^{\infty} \left[ \frac{-1}{x^2}e^{-x^2y} \right]_{0}^{\infty}dx\\ &=\int_{1}^{\infty} \frac{1}{x^2}dx\\ &=\left[ \frac{-1}{x} \right]_{1}^{\infty}\\ &=1. \end{align*}
Therefore, we know that $f_{X,Y}$ is a density function and we can move on to the next question. It is asked to evaluate the folowing probabilty : $\mathbb{P}(X^2Y>1)$.
I am unsure how to find the right boundaries for the integrals. Here's what I've done so far :
The condition $X^2Y>1$ gives us the following inequalities :
$$1 \leq x < \infty \hspace{1cm} \text{and} \hspace{1cm} \frac{1}{x^2} \leq y < \infty$$
or
$$0 \leq y < \infty \hspace{1cm} \text{and} \hspace{1cm} \frac{1}{\sqrt{y}} \leq x < \infty.$$
From that I deduce the two folowing double integrals :
$$ \mathbb{P}(X^2Y>1)=\int_{1}^{\infty} \left( \int_{x^{-2}}^{\infty} e^{-x^2y}dy \right)dx $$
and
$$ \mathbb{P}(X^2Y>1)=\int_{0}^{\infty} \left( \int_{y^{-1/2}}^{\infty} e^{-x^2y}dx \right)dy. $$
I already evaluated the first integral but the second one is more complicated. I am right ? How do you find the boundaries for such problems ?
Thanks
The first integral is correct and gives you the right solution:
$$\mathbb{P}[X^2Y>1]=\frac{1}{e}$$
the second one is wrong
The correct one should be
$$\mathbb{P}[X^2Y>1]=\int_0^{1}\Bigg[\int_{y^{-\frac{1}{2}}}^{\infty}e^{-x^2y}dx \Bigg]dy+\int_1^{\infty}\Bigg[\int_{1}^{\infty}e^{-x^2y}dx \Bigg]dy$$
It is self evident that it is better the first choice.
This is a clear example showing that not always integration's order is neutral.
The requested probability is
$$\mathbb{P}[X^2Y>1]=\mathbb{P}[Y>\frac{1}{X^2}]$$
Thus the area to be integrated is the following purple one (the blu curve is the equation $y=\frac{1}{x^2}$)
In your second way to solve it you tried to integrate also the yellow one