1. Let Z be the time it takes from zero for a meteor of size x or larger to strike the moon.

We will now calculate the probability of P (Z ≥ z ) , where z is some positive number representing time in units congruent to that of the arrival rate:

Let N Po(λz), i.e. it is the number of meteors that will strike the moon in time period of length z.

Also, let X represent the size of meteors. So, X has the CDF GX (.).

So,

P (Z ≥ z )=P (N=0 )+P (N=1 )GX ( x )+P (N=2 )[GX (x )]2+P (N=3 )[GX (x )]3+…=∑i=0

∞

P (N=i )[GX ( x )]i

.

(Explanation: for a meteor of size more than x to hit the moon after time z has passed, one of the following mutually exclusive cases must be true:

i. 0 meteor hit the moon in time z.ii. 1 meteor hits the moon in time z but its size smaller than x (the probability of which isGX ( x )).iii. 2 meteors hit the moon in time z but both are smaller than x (the probability of which is[GX ( x )]2 because of independence).

iv. 3 meteors hit the moon in time z but all are smaller than x (the probability of which is[GX ( x )]3 because of independence).v. and so on…)

P (Z ≥z )=∑i=0

∞

P (N=i )[GX (x )]i=∑i=0

∞

e− λz( λz)i

i ! [GX ( x )]i=e−λz∑i=0

∞ ( λzGX ( x ))i

i ! =e− λz eλz GX ( x )

(Explanation: ∑i=0

∞ ( λzGX ( x ))i

i ! =eλzGX ( x )by Taylor series of the exponential function)

P (Z ≥ z )=e−λze λzGX (x )=e− λz(1−GX (x )), the survivor function of Z. Survivor functions of non-negative random variables have the following property:

E (Z )=∫0

∞

P (Z≥ z )dz=∫0

∞

e− λz(1−GX (x ))dz=[ e− λz (1−G X ( x ))

−λ (1−GX ( x ) ) ]0∞

= 1λ (1−GX (x ) )

.

(Explanation: this answer also makes sense. If x=0, i.e. the time to for meteor of any size to strike

the moon was the case, GX (0 )=0 making E(Z )=1λ (which is the time to the first occurrence of a

Poisson process).