Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 5

The cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

The continuous nonnegative part has total mass ${\lim }_{y\to \infty }(1-e^{-y})/2=1/2$. The remaining mass is at $-1$, giving $\boxed{m=1/2}$. This is a mixed probability distribution, specifically a mixture of a point mass and an exponential distribution; its atom must be included in normalization.

Include the negative atom as well as the continuous part in the expected value:

The second moment of the mixed probability distribution is $\mathbb{E}[Y^2]=\frac{1}{2}(-1{)}^2+\frac{1}{2}{\int }_0^{\infty }{y}^2{e}^{-y}dy=1/2+1=3/2$. Since the mean is zero,

The conditioning event selects the exponential component: $P(Y\ge 0)=1/2$ and $P(Y\ge 1)=e^{-1}/2$. Thus conditional probability givesThe atom at $-1$ is excluded by the conditioning event.