Joint, marginal and conditional distributions

2 free worked examples

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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. $P={\int }_0^1{\int }_0^{1-x}(x+y)dydx$.
   3. The inner integral is $x(1-x)+\frac{(1-x{)}^2}{2}$.
   4. Integrating over $x$ from 0 to 1 gives $\frac{1}{2}-\frac{1}{3}+\frac{1}{6}=\frac{1}{3}=0.3333$.

   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 $\mathrm{Var}$ 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: $\mathrm{Var}(S)=E\left[\mathrm{Var}(S\mid N)\right]+\mathrm{Var}\left(E[S\mid N]\right)$.
   2. $E\left[\mathrm{Var}(S\mid N)\right]=E[10,000N]=10,000(4)=40,000$.
   3. $\mathrm{Var}\left(E[S\mid N]\right)=\mathrm{Var}(300N)=90,000\mathrm{Var}(N)=90,000(4)=360,000$.
   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.