MTH5121 Probability Models
Solutions to Exercise Sheet 5
1. In each of the following cases find the probability that the branching process
generated by X has died out at time n for 1 ≤ n ≤ 5 and find the probability
that it dies out at some time (i.e., that it goes extinct).
a) X has distribution P(X = 0) = 1/2, P(X = 1) = 1/4 and P(X = 2) = 1/4.
b) X has distribution P(X = 0) = 1/4, P(X = 1) = 1/4 and P(X = 2) = 1/2.
c) X has distribution P(X = 0) = 1/4, P(X = 1) = 1/2 and P(X = 2) = 1/4.
Solution: Let Yn be the number of cells at time n, let θn be the probability
Yn = 0 and let GX be the probability generating function of X. We proved in
lectures that θ0 = 0 and θn = GX (θn−1 ) for n > 0. (Explicitly θ1 = GX (θ0 )
and θ2 = GX (θ1 ) etc.)
a) GX (t) = 1/2 + t/4 + t2 /4. Thus we find that θ1 , θ2 , . . . , θ5 are 0.5, 0.69,
0.79, 0.85, 0.90.
The probability of extinction is 1 since E(X) = 3/4 < 1.
Alternatively the probability of extinction is the least non-negative root
of GX (t) = t: i.e.,of 1/2+t/4+t2 /4 = t which rearranges to t2 −3t+2 = 0.
This has roots t = 2 and t = 1: therefore 1 is the least non-negative root
and the probability of extinction is 1.
b) This time GX (t) = 1/4 + t/4 + t2 /2 and we get probabilities 0.25, 0.35,
0.40, 0.43, 0.45. The probability of extinction is the least positive root of
the equation 1/4 + t/4 + t2 /2 = t i.e., of 2t2 − 3t + 1 = 0. This has roots
t = 1 and t = 1/2, so the probability of extinction is 1/2.
c) This time GX (t) = 1/4 + t/2 + t2 /4 and we get probabilities 0.25, 0.39,
0.48, 0.55, 0.60.
Since E(X) = 1 (and X is not always 1) we see that the probability of
extinction is 1.
Alternatively the probability of extinction is the least non-negative root
of GX (t) = t: i.e.,of 1/4+t/2+t2 /4 = t which rearranges to t2 −2t+1 = 0.
This has a repeated root at t = 1: therefore 1 is the least non-negative
root and the probability of extinction is 1.
Comment: The alternatives are just that: either solution is fine.
Do note what least non-negative solution means: it means find all solutions
and pick the smallest one which is not negative.
Do note how easy the calculations are: the theory was quite technical but
having proved the results the calculations are easy.
2. Suppose that X is a random variable and that GX is its probability generating
function.
a) What is GX (1)?
b) Using part (a) state a solution to the equation GX (t) = t.
c) Using part (b) state one factor of the equation Gx (t) − t = 0.
Now suppose that GX (t) = (t3 + t2 + t + 1)/4.
d) Use part (c) to find all solutions to the equation GX (t) − t = 0.
e) What is the probability of extinction for the branching process generated
by X?
f) State a distribution X which has probability generating function (t3 +t2 +t+1)/4.
Solution:
a) GX (t) =
P∞
n=0 P(X
= n)tn so
GX (1) =
∞
X
n=0
n
P(X = n)1 =
∞
X
P(X = n) = 1.
n=0
b) Hence t = 1 is a solution to the equation GX (t) = t.
c) Hence t − 1 is a factor of the equation GX (t) − t = 0.
d) We need to solve GX (t)−t = 0 which is the same as (t3 +t2 +t+1)/4−t = 0
which rearranges to
t3 + t2 − 3t + 1 = 0.
Now we know that t − 1 is a factor. So we can remove this factor (using
long division for example) to get
(t − 1)(t2 + 2t − 1) = 0.
Hence either t − 1 = 0 (so t = 1), or t2 + 2t − 1 = 0. The latter is a
quadratic
√ which we√can solve using the quadratic formula to find that
−2± 8
t=
= −1 ± 2.
2
√
Thus there are three solutions to GX (t) = t: namely, t = 1, t = −1 − 2,
√
and t = −1 + 2.
e) The probability of extinction of the branching process generated by X
is the least non-negative solution to GX (t) = t which is the same as the
equation GX (t) − t = 0 we have just solved.
√
√
Obviously t = −1 − 2 is negative. Also t = −1 + 2 is between 0 and
√
1 (since 2 is between 1 and 2). Hence the least non-negative solution is
√
t = −1 + 2. (If you put it in a calculator you find t = 0.414).
In other words the extinction probability is t =
√
−2+ 8
.
2
f) We could use the formula for the mass function but it is much easier than
that. Suppose X has distribution given by
P(X = 3) = P(X = 2) = P(X = 1) = P(X = 0) = 1/4.
Then X would have pgf GX .
More generally if we have the pgf written as a polynomial (or power series)
then we can read off the mass function: it is just the coefficients of the
powers of t. That is the P(X = n) is the coefficient of tn .
3. Suppose that Y0 , Y1 , Y2 . . . is the branching process generated by a distribution
X with probability generating function GX and mean µ.
a) What are GX (1) and G0X (1)? [Justification is not required]
b) Using the fact that
GYn (t) = GYn−1 (GX (t))
prove that E(Yn ) = E(Yn−1 )µ.
Solution:
a) GX (1) = 1 ( this is true for any probability generating function) and
G0X (1) = E(X) = µ.
b) We differentiate this formula to see that:
G0Yn (t) = G0Yn−1 (GX (t))G0X (t).
We know that G0Yn (1) = E(Yn ) and G0Yn−1 (1) = E(Yn−1 ). Hence
E(Yn ) = G0Yn (1) = G0Yn−1 (GX (1))G0X (1)
= G0Yn−1 (1)µ
by part (a)
= E(Yn−1 )µ.
4. Suppose that Y0 , Y1 , Y2 . . . is the branching process generated by a distribution
X with mean µ.
a) Find E(Yn+1 | Yn = λ) and E(Yn+1 Yn | Yn = λ).
b) What is E(Yn+1 Yn | Yn )?
c) Use your answer to part (b) to show that
E(Yn+1 Yn ) = µE(Yn2 ).
Solution:
a) We know from lectures that E(Yn+1 | Yn = λ) = λE(X) = λµ. Or we can
prove this directly
λ
λ
X
X
E(Yn+1 | Yn = λ) = E(
Xi ) =
E(Xi ) = λE(X) = λµ
i=1
i=1
where the Xi are independent copies of X.
Now
E(Yn+1 Yn | Yn = λ) = E(λYn+1 | Yn = λ) = λE(Yn+1 | Yn = λ) = λ2 µ
(where the first step follows since we are conditioning on Yn = λ).
b) Thus E(Yn+1 Yn | Yn ) = Yn2 µ.
c) We apply the tower law for expectation:
E(Yn+1 Yn ) = E(E(Yn+1 Yn | Yn )) = E(Yn2 µ) = µE(Yn2 )
as required.