A tricky joint probability density problem

\(\)Here is problem 7.1.9 from my current probability & statistics textbook, Probability and Statistical Inference by Bartoszynski and Bugaj, which I’m using in the Master’s-level Statistical Theory class taught by Dr. Bugaj herself at Western Michigan University:

Variables \(X\) and \(Y\) have the joint density \(f(x,y) = 1/y\) for \(0 < x < y < 1\) and \(f(x,y) = 0\) otherwise. Find \(P(X + Y > 0.5)\).

As with most joint pdf/cdf problems, the two most important and difficult things to do are identify the correct region over which to integrate and define the bounds of integration correctly. I made a basic graph of the region that we need to integrate over:

I don’t know how to shade a region, but the region we need is above the two graphed lines and below \(y=1\). The line with the positive slope is \(y=x\), and the line with the negative slope is \(x+y=0.5\), or \(y=-x+\frac{1}{2}\). The given pdf is only valid for \(0< x < y < 1\), i.e., when \(y > x\) but \(y < 1\), so that's why the line \(y=x\) is needed regardless of the specific question. And this specific question asks for the probability that \(X+Y>0.5\), i.e., when \(Y>0.5-X\), so we need to consider only the part of the pdf above the line \(y = -x + \frac{1}{2}\). Therefore, the region we need to integrate over is the quadrilateral with vertices \((0, \frac{1}{2}), (0, 1), (1, 1)\), and \((\frac{1}{4}, \frac{1}{4})\).

Whether we put the \(dx\) or \(dy\) on the inside or outside of the double integral, we have to break this region up into two sub-regions—at least, that’s the only way I know of to integrate this type of region. I’ll first show how to integrate this with \(dx\) on the outside and \(dy\) on the inside, and then vice versa.

(i) To integrate with the \(x\)-direction on the outside and the \(y\)-direction on the inside, the two sub-regions of this quadrilateral have to be from \(x=0\) to \(x=\frac{1}{4}\) and from \(x=\frac{1}{4}\) to \(x=1\). This makes the vertical bounds of integration \(y=-x+\frac{1}{2}\) to \(y=1\) and \(y=x\) to \(y=1\), respectively.

Thus,
$$
\begin{eqnarray}
P(X+Y>0.5) &=& \int_0^.25 \int_{0.5-x}^1 \frac{1}{y}\,dy\,dx + \int_{0.25}^1 \int_x^1 \frac{1}{y}\,dy\,dx \nonumber \\[9pt]
&=& \int_0^{0.25} -\ln (0.5-x)\,dx + \int_{0.25}^1 -\ln x\,dx \nonumber \\[9pt]
&=& (0.25) + (0.75 + 0.25\ln 0.25) \nonumber \\[9pt]
&=& 1 + 0.25\ln 0.25 \nonumber \\[9pt]
&\approx& 0.6534 \\
\end{eqnarray}
$$

You can do the left-hand integral using integration by parts, but in this case, it conveniently computes to \(0.25\), which I found by using my TI-83 or Wolfram Alpha.

(ii) To integrate with the \(y\)-direction on the outside and the \(x\)-direction on the inside, we have to divide the region into two sub-regions vertically. The bottom sub-region goes from \(x = 0.5 – y\) to \(x = y\) and from \(y = \frac{1}{4}\) to \(y = \frac{1}{2}\), and the top sub-region goes from \(x = 0\) to \(x = y\) and from \(y = \frac{1}{2}\) to \(y = 1\).

$$
\begin{eqnarray}
P(X+Y>0.5) &=& \int_{0.25}^{0.5} \int_{0.5-y}^y \frac{1}{y} \,dx\,dy + \int_{0.5}^1 \int_0^y \frac{1}{y}\,dx\,dy \nonumber \\[9pt]
&=& \int_{0.25}^{0.5} \left(\frac{y}{y} – \frac{0.5-y}{y}\right)\,dy + \int_{0.5}^1 1 \,dy \nonumber \\[9pt]
&=& \int_{0.25}^{0.5} \left(2 – \frac{1}{2y}\right) \,dy + \int_{0.5}^1 1 \,dy \nonumber \\[9pt]
&=& (0.5 + 0.5\ln 0.25 – 0.5\ln 0.5) + (0.5) \nonumber \\[9pt]
&=& 1 + 0.5\ln 0.25 – 0.5\ln 0.5 \nonumber \\[9pt]
&\approx& 0.6534
\end{eqnarray}
$$

Looking at the second-to-last line of each of these two equation displays, we can see that \(1 + 0.25\ln 0.25\) and \(1 + 0.5\ln 0.25 – 0.5\ln 0.5\) add to the same number. Therefore,
$$
\begin{eqnarray}
1 + 0.25\ln 0.25 &=& 1 + 0.5\ln 0.25 – 0.5\ln 0.5\nonumber \\[9pt]
0.25\ln 0.25 &=& 0.5\ln 0.25 – 0.5\ln 0.5\nonumber \\[9pt]
0.5\ln 0.5 &=& 0.25\ln 0.25
\end{eqnarray}
$$

How is that?! It results from the “logarithm power rule”: \(\log_a (b^y) = y\log_a b\). So with those natural logarithms above, we can bring the coefficient up as an exponent of the argument of the \(\ln\):
$$
\begin{eqnarray}
0.5\ln 0.5 &=& 0.25\ln 0.25\nonumber \\[9pt]
\ln (0.5^{0.5}) &=& \ln (0.25^{0.25}) \nonumber \\[9pt]
e^{\ln (0.5^{0.5})} &=& e^{\ln (0.25^{0.25})} \nonumber \\[9pt]
0.5^{0.5} &=& 0.25^{0.25} \nonumber \\[9pt]
\frac{1}{2^{\frac{1}{2}}} &=& \frac{1}{4^{\frac{1}{4}}} \nonumber \\[9pt]
\frac{1}{\sqrt{2}} &=& \frac{1}{(2^2)^{\frac{1}{4}}} \nonumber \\[9pt]
\frac{1}{\sqrt{2}} &=& \frac{1}{2^{\frac{1}{2}}} \nonumber \\[9pt]
\frac{1}{\sqrt{2}} &=& \frac{1}{\sqrt{2}}
\end{eqnarray}
$$

Finally, I thought it was worth noting that the integration approach that seemed the most natural to me, dividing the region into two sub-regions vertically (i.e., number ii above), actually involved slightly harder algebra and a longer expression for the answer. Ultimately, though, it’s always best to choose the integration approach that you’re most comfortable with and confident in, unless you know the integral is going to be much harder that way for some reason (integration by parts, for example).

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