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

The continuous nonnegative part has total mass

lim_{y \to \infty} (1 - e^{- y}) /2 = 1/2

. The remaining mass is at

- 1

m = 1/2

ii

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

E [Y] = \frac{1}{2} (- 1) + \frac{1}{2} \int_{0}^{\infty} y e^{- y} d y = - \frac{1}{2} + \frac{1}{2} = 0.

E [Y^{2}] = \frac{1}{2} (- 1)^{2} + \frac{1}{2} \int_{0}^{\infty} y^{2} e^{- y} d y = 1/2 + 1 = 3/2

. Since the mean is zero,

Var (Y) = \frac{3}{2} .

The conditioning event selects the exponential component:

P (Y \geq 0) = 1/2

and

P (Y \geq 1) = e^{- 1} /2

P (Y \geq 1 ∣ Y \geq 0) = \frac{e ^{- 1} /2}{1/2} = e^{- 1} .

The atom at

- 1

is excluded by the conditioning event.

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