Finding probability a loss will be more than twice the second loss Question: Losses follow an exponential distribution with mean $1$. Two independent losses are
observed.
Calculate the probability that either of the losses is more than twice the other.
My Attempt: $P(L_1 < x) = 1-e^{-x}$, $P(L_2 > 2x) = e^{-2x}$ <-- This step is where I go wrong. Instead of $P(L_1 < x)$ apparently I'm supposed to use $P(L_1 > x)$ but for the life of me I can't conceptualize why. Doesn't this allow for loss $1$ to be $3$ and loss $2$ to be $4$?
$$P(L_1 < x \land L_2 > 2x) = e^{-2x}-e^{-3x}$$
Integrating over all $x$: $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
Multiplying by $2$ to account for Loss $2$ being bigger than Loss $1$: $\frac{1}{6}\times2 = \frac{1}{3}$
My answer: $\frac{1}{3}$ Correct answer: $\frac{2}{3}$
 A: There's a few problems with your proof:

*

*By considering $P(L_1<x, L_2>2x)$, you're not accounting for some possibilities, e.g. if $x=4$, this doesn't allow $L_1=3$, $L_2=7$. What you want is $P(L_2>2L_1)$

*Integrating $P(L_1<x, L_2>2x)$ over all $x$ makes no sense. Remember you only ever integrate PDFs.

With that said, here's how to do it:

*

*The joint PDF is $f(x,y)=e^{-x}e^{-y}$ because $L_1, L_2$ are independent.

*As I mentioned in 1 above, what we're interested in is $P(L_2>2L_1)=\int_{x=0}^\infty \int_{y=2x}^\infty f(x,y) \; dx dy = 1/3$

*As you correctly noticed, you then need to multiply this by 2 to account for the other way round

A: For problems like these, I encourage you to think about the losses as represented by an ordered pair, say $(X, Y)$, and plotted on a coordinate plane.  So $X$ and $Y$ are independent and identically distributed exponential random variables with densities $f_X(x) = e^{-x}$ and $f_Y(y) = e^{-y}$.
Now consider the event that neither loss is twice the other (i.e., the complementary event of what is being asked in the question).  In the first quadrant of the $XY$-coordinate plane, imagine coloring all ordered pairs $(X,Y)$ that meet this condition.  What does this region look like?
A little thought (and maybe some sketching) should convince you that the region is bounded by two lines, one corresponding to $Y = 2X$, and (by an elementary symmetry argument) the other by $X = 2Y$.  Thus we can characterize this region by the compound inequality
$$0 < X/2 < Y < 2X. \tag{1}$$
It follows that the desired probability is simply $1 - \Pr[X/2 < Y < 2X]$, remembering that the event $(1)$ is the complement of what we wish to compute.  How do we do the calculation?  One way is to simply compute the double integral of the joint density of $(X,Y)$ over the specified region:
$$\begin{align}
\Pr[X/2 < Y < 2X]
&= \int_{x=0}^\infty \int_{y=x/2}^{2x} f_{X,Y}(x,y) \, dy \, dx \\
&= \int_{x=0}^\infty \int_{y=x/2}^{2x} e^{-x} e^{-y} \, dy \, dx \\
&= \int_{x=0}^\infty e^{-x} \bigl[ -e^{-y} \bigr]_{y=x/2}^{2x} \, dx \\
&= \int_{x=0}^\infty e^{-x} (e^{-x/2} - e^{-2x}) \, dx \\
&= \int_{x=0}^\infty e^{-3x/2} - e^{-3x} \, dx \\
&= \left[ -\frac{2}{3}e^{-3x/2} + \frac{1}{3}e^{-3x} \right]_{x=0}^\infty \\
&= 0 - \left( -\frac{2}{3} + \frac{1}{3} \right) \\
&= \frac{1}{3}. \tag{2}
\end{align}$$
Then again remembering to compute the complement, the desired probability is $1 - 1/3 = 2/3$, as claimed.
Another way to solve the question, and probably the originally intended method, is to note the symmetry of the variables $X$ and $Y$ and compute directly the probability that, say, the second loss is more than twice the first; e.g.,
$$\Pr[Y > 2X] = \int_{x=0}^\infty S_Y(2x) f_X(x) \, dx, \tag{3}$$ where $S_Y = 1 - F_Y$ is the survival function.  But since $S_Y$ is the same as $S_X$, as $X$ and $Y$ are iid, then we have $$\Pr[Y > 2X] = \int_{x=0}^\infty e^{-2x} e^{-x} \, dx = \int_{x=0}^\infty e^{-3x} \, dx = \frac{1}{3}. \tag{4}$$
But now you have to remember that this is only one half of the total probability, because by symmetry, the other event is $X > 2Y$--the first loss is more than twice the second.  So the answer is $2/3$, which is what we obtained after taking the complement of the double integral in $(2)$.

As a further exercise, how would you generalize this problem:  suppose $X$ and $Y$ remain independent exponentially distributed random variables, but are no longer identically distributed?  Specifically, say $X$ has mean $\mu$ and $Y$ has mean $\theta$.  And instead of twice, consider the outcome where one loss is more than $k$ times the other, for some $k > 1$.  What is the desired probability in terms of $k, \mu, \theta$?  Which computational approach would you prefer using?
