Joint Density Function Example f(x,y)= $\frac 1y$  for 0 &lt; x &lt; y &lt;1

f(x,y)= 0 elsewhere

find P(x+y>.5):

$\iint$ $\frac 1y$ dxdy

I basically can't figure out the limits, since x and y are not only dependent on each other but also the probability statement itself.
$\displaystyle{%
{\rm f}\left(x,y\right)
=
{\Theta\left(x)\Theta\left(y - x\right)\Theta(1 - y\right) \over y}}$


\begin{align}
&amp;\color{#ff0000}{\large\int_{-\infty}^{\infty}{\rm d}x
\int_{-\infty}^{\infty}{\rm d}y\,{\rm f}\left(x,y\right)}
=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}
\int_{-\infty}^{\infty}{\rm d}x\,
\Theta\left(x\right)\Theta\left(y - x\right)
\\[3mm]&amp;=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}
\Theta\left(y\right)\int_{0}^{y}{\rm d}x\,
=
\int_{-\infty}^{\infty}{\rm d}y\,\Theta\left(1 - y\right)\Theta\left(y\right)
=
\int_{0}^{1}{\rm d}y = 
\color{#ff0000}{\large 1}
\end{align}



\begin{align}
\color{#ff0000}{\large{\rm P}\left(x + y &gt; 5\right)}
&amp;=
\int_{-\infty}^{\infty}{\rm d}x
\int_{-\infty}^{\infty}{\rm d}y\,{\rm f}\left(x,y\right)
\Theta\left(x + y - 5\right)
\\[3mm]&amp;=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}\,
\Theta\left(y\right)
\int_{0}^{y}{\rm d}x\,\Theta\left(x + y -5\right)
\\[3mm]&amp;=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}\,
\Theta\left(y\right)
\overbrace{\int_{y - 5}^{2y - 5}{\rm d}x\,\Theta\left(x\right)}
^{\left(~=\ 0\ \mbox{since}\ y - 5\ &lt;\ 2y - 5\ &lt;\ 0\
         \mbox{when}\ 0\ &lt;\ y\ &lt;\ 1~\right)}
=
\color{#ff0000}{\large 0}
\end{align}

This result is quite obvious since ${\rm P}\left(x &gt; 1\right) = 0$
and ${\rm P}\left(y &gt; 1\right) = 0$ such that ${\rm P}\left(x + y &gt; 2\right) = 0$.


In general
$$
\left\langle\phi\right\rangle
\equiv
\int_{0}^{1}{{\rm d} y \over y}\int_{0}^{y}{\rm d}x\,\phi\left(x,y\right)
$$
The region given is $ 0 &lt; x&lt; y&lt;1$. In particular, this gives $0&lt;x&lt;y$

so we have $\int\int f(x,y) dx dy = \int_0^1\int_0^y \frac{1}{y} dx dy + \int_0^1\int_y^1 0 dx dy$

We can evaluate this, and we get

$\int_0^1\int_0^y \frac{1}{y} dx dy + \int_0^1\int_y^1 0 dx dy = \int^1_0 \frac{1}{y} *(y-0)dy + 0 = \int_0^1 1 dy = 1$

(edit) So to answer the probability statement for $P(x+y&gt;0.5)$

$x+y&gt;\frac{1}{2} \Leftrightarrow x &gt; \frac{1}{2}-y$, so we are interested in the intersection of the regions $0&lt;x&lt;y&lt;1$ and $1/2-y &lt; x&lt;1$. Combined, these give the region $\frac{1}{2} -y &lt; x&lt; y$ and $1/2 &lt; y&lt;1$

$$\int_0^1 \int_{1/2-y}^1 f(x,y) dx dy = \int_.5^1\int_{1/2-y}^y \frac{1}{y}dxdy$$

$$= \int_.5^1 \frac{1}{y}(2y-1/2) = \int_.5^1 2-\frac{1}{2y}$$
$$= 2*.5 -\frac{1}{2}(\ln(1) - \ln(.5))$$
$$= 1 - \ln(\sqrt{2})$$