Joint, marginal and conditional distributions

2 free worked examples

Solved in full, free to everyone, and never rotated. Read these before deciding whether the rest of the drill is worth paying for.

1. Exam PJoint, marginal and conditional distributionsExam level

   Two standardised losses have joint density $f(x,y) = x+y$ on the unit square. Find $P(X+Y<1)$.

   1. A$0.2500$
   2. B$0.3333$
   3. C$0.5000$
   4. D$0.6667$
   5. E$0.7500$

   Solution

   1. Sketch the region: the triangle below the line $y = 1-x$ inside the unit square.
   2. $P = \int_0^1\int_0^{1-x}(x+y)\,dy\,dx$.
   3. The inner integral is $x(1-x) + \dfrac{(1-x)^{2}}{2}$.
   4. Integrating over $x$ from 0 to 1 gives $\tfrac12 - \tfrac13 + \tfrac16 = \tfrac13 = 0.3333$.

   Trap. Using constant limits 0 to 1 for y, which integrates over the whole square rather than the triangle.

2. Exam PJoint, marginal and conditional distributionsStretch

   $N$ is Poisson with mean 4, and given $N = n$ the claim total has mean $300n$ and variance $10{,}000n$. Find $\operatorname{Var}$ of the total.

   1. A$40{,}000$
   2. B$360{,}000$
   3. C$400{,}000$
   4. D$1{,}200{,}000$
   5. E$1{,}440{,}000$

   Solution

   1. Use the conditional variance decomposition: $\operatorname{Var}(S) = E\left[\operatorname{Var}(S\mid N)\right] + \operatorname{Var}\left(E[S\mid N]\right)$.
   2. $E\left[\operatorname{Var}(S\mid N)\right] = E[10{,}000N] = 10{,}000(4) = 40{,}000$.
   3. $\operatorname{Var}\left(E[S\mid N]\right) = \operatorname{Var}(300N) = 90{,}000\operatorname{Var}(N) = 90{,}000(4) = 360{,}000$.
   4. Total: $40{,}000 + 360{,}000 = 400{,}000$.

   Trap. Using only one of the two terms - the 'variance of the mean' piece is the one usually forgotten.