Probability 1 Exercises of week 5

Probability 1
Exercises of week 5
Please solve the starred problems, the others will be discussed in the exercise classes.
Please hand in your solutions before 25/03/2014.
Exercise 1. (i) In a gamble, the probability to win is p ∈ [0, 1] and the probability to lose is
q = 1 − p. A gambler played the game for n ≥ 3 times. Calculate the probability that the
gambler
(a) won at least 1 time;
(b) lost at least (n − 2) times;
(c) won at least 1 time and lost at least 1 time.
(ii) For an arbitrary n ≥ 3, let Xn be a random variable on {1, 2, · · · , n − 1}, whose distribution
is
P [Xn = k] = Cn max(k, n − k), k = 1, 2, · · · , n − 1.
Determinate the value of constant Cn ; calculate E [Xn ] and E [n − Xn ].
Exercise 2.
(i) Suppose that a probability measure has density f :
f (x) = C(1 − exp(−
1
)), ∀x ∈ R.
x2
Determinate the value of constant C.
√
R
u2
Hint: R e− 2 du = 2π.
(ii) (standard Cauchy distribution) Suppose that a probability measure has density f :
f (x) = C
1
, ∀x ∈ R.
1 + x2
Determinate the value of constant C. If the random variable X has density f , prove that the
expected value of X is infinite.
Exercise 3. For θ ∈ (0, 21 ), n ∈ N, consider a sequence un defined by
Z θ n+1
x
un =
dx
2
0 1−x
Calculate limn→+∞ un ; then give a constant C, such that (pn := Cun )n∈N defines a probability on
N.
*Exercise 4. For θ ∈ (0, 1) and n ≥ 2, define a sequence (pk , k = 1 · · · n) by
Cn (1 − θ) min(k, n − k), if k = 1, · · · , n − 1,
pk =
θ
if k = n.
Determinate the value of constant Cn , such that (pk , k = 1 · · · n) is a probability measure on
{1, · · · , n}.
*Exercise 5. Recall the definition of random walk Sn : we set S0 = 0, and Sn = X1 + · · · + Xn ,
∀n ≥ 1. For any i ∈ N, P [Xi = 1] = P [Xi = −1] = 12 .
(i) For r ≥ 1, calculate the number of paths from time 0 and 2n−1, such that S0 = 1, S2n−1 = 2r,
and the path touches or crosses the t-axis at sometime between 0 and 2n − 1.
(ii) Calculate P [S1 > 0, · · · , S2n−1 > 0, S2n = 2r] for r ≥ 1.
(iii) Prove that
P [S1 6= 0, · · · , S2n 6= 0] = 2P [S1 > 0, · · · , S2n > 0] = P [S2n = 0] .
2