17.1SB2 Random variables & their distributions Tutorial Examples III

17.1SB2 Random variables & their distributions Tutorial Examples III

→ 1. In each of the following, a probability function f(x) is given. Find the mean and variance in each case.

(a)

x : 0 1 2 3 4

fX(x) : 1/15 2/15 3/15 4/15 5/15 (b)

x : −2 −1 0 1 2

f_X(x) : 1/16 4/16 6/16 4/16 1/16 (c)

x : 1 2 3 4 5

f_X(x) : 1/16 4/16 6/16 4/16 1/16 Solution:

(a) E(X) = 0× 1/15 + 1 × 2/15 + 2 × 3/15 + 3 × 4/15 + 4 × 5/15 = 8/3

and E(X²) = 0× 1/15 + 1 × 2/15 + 4 × 3/15 + 9 × 4/15 + 16 × 5/15 = 26/3. Hence Var(X) = E(X²)− E(X)²= 26/3− 64/9 = 14/9.

(b) E(X) = 0 and E(X²) = 1, so Var(X) = 1.

(c) E(X) = 3 and E(X²) = 10, so Var(X) = 1. [Alternatively, use the fact that X = Y + 3 where X is a random variable having probability function given by (c) and Y is a random variable whose probability function is given by (b).]

→ 2. A coin is given 5 independent tosses, with probability p of landing Heads and probability q = 1− p of landing Tails at each toss. Let the random variables H and T be the total number of Heads and Tails respectively.

(a) Write down the probability function of H and find E(H) and Var(H).

(b) By defining suitable indicator random variables (see Tutorial Sheet II), use an alternative simpler way to find E(H).

(c) (Harder!) Use the method of (b) to find Var(H).

Solution:

(a) The probability function of H is fH(x) =

5 x



p^x(1− p)^5−x.

(Every outcome containing x Heads and 5− x Tails has probability p^x(1− p)^5−x, and there are C_x⁵ such outcomes because that’s how many ways are there of deciding where the Heads are.) Elementary, but rather messy algebra using the straightfor- ward definition of E(H) and Var(H) give E(H) = 5p (which should be obvious!) and Var(H) = 5p(1− p).

(b) Define random variables I_j, j = 1, 2, . . . , 5 to be I_j =

1 if j^th toss lands Heads 0 otherwise.

Then H = I1 + I2 +· · · + I5. Now E(Ij) = P (Ij = 1) = p for all j, so E(H) = E(I1) +· · · + E(I5) = 5p.

(c) Define the I_j’s as in (b). Then Var(I_j) = E(I_j²)−E(Ij)² = E(I_j)−E(Ij)² = p−p². (Note that I_j² = I_j.) Since the coin tosses are independent, the random variables Ij and Ik (j 6= k) are independent random variables. Hence it is valid to add the individual variances of the Ij’s in the following calculation:

Var(X) = Var(I₁+ I₂+· · · + I5) = Var(I₁) +· · · + Var(I5) = 5(p− p²) = 5p(1− p).

→ 3. Let X have Poisson distribution with mean 3. In each of the following, use the formula for the probability function of the Poisson distribution to find the answer, then compare your answer with that obtained by looking up statistical tables.

(a) P (X ≤ 3); (b) P (X = 0);

(c) P (X > 1); (d) P (1≤ X ≤ 4).

Solution:

(a)

P (X ≤ 3) = X3 k=0

e⁻³3^k

k! = 0.6472.

(b) P (X = 1) = 3e⁻³ = 0.149361. Alternatively, if using tables P (X = 1) = P (X ≤ 1)− P (X ≤ 0) = 0.1991 − 0.0498 = 0.1493.

(c) P (X > 1) = 1− P (X ≤ 1) = 1 − 0.1991 = 0.8009.

(d)

P (1≤ X ≤ 4) = X4 k=1

e⁻³3^k

k! (direct method)

= P (X ≤ 4) − P (X ≤ 0) = 0.8153 − 0.0498 (using tables)

= 0.7655.

→ 4. Let X have Poisson distribution with mean 40. Use the normal approximation to find

(a) P (30≤ X ≤ 45); (b) P (25≤ X ≤ 55);

(c) P (X = 40); (d) P (X ≥ 35).

Solution:

Let

Z = X√− 40 40 .

Then the normal approximation says that Z is approximately a N (0, 1) random variable.

(a)

P (30≤ X ≤ 45) = P ((30 − 40)/√

40≤ Z ≤ (45 − 40)/√ 40)

= P (−1.581139 ≤ Z ≤ 0.790569)

≈ Φ(0.790569) − Φ(−1.581139)

= Φ(0.79)− (1 − Φ(1.58)) = 0.7852 − 1 + 0.9429 = 0.7281.

(b) Similarly, P (25≤ X ≤ 55) ≈ 0.98222.

(c)

P (X = 40) = P (X ≤ 40) − P (X ≤ 39)

= P (Z ≤ (40 − 40)/√

40)− P (Z ≤ (39 − 40)/√ 40)

≈ Φ(0) − Φ(−0.158)

= Φ(0)− (1 − Φ(0.158)) = 0.5 − 1 + 0.5636 = 0.0636.

(d)

P (X ≥ 35) = 1 − P (X < 35)

= 1− P (Z < −0.79) ≈ 1 − Φ(−0.79)

= 1− (1 − Φ(0.79)) = Φ(0.79) = 0.7852.

5. The number of particles emitted by a radioactive source during a fixed period is a random variable with Poisson distribution of mean λ:

fN(x) = e^−λλ^x

x!, x = 0, 1, 2, 3, . . . Find E(N ) and E(N²).

Hint: Use

X∞ x=0

fN(x) = X∞ x=0

e^−λλ^x x! = 1.

Solution:

E(N ) = X∞ x=0

xfN(x) = e^−λ X∞ x=0

xλ^x x!

= e^−λ X∞ x=1

xλ^x

x! (the x = 0 contributes nothing, so can sum from x = 1)

= e^−λ X∞ x=1

λ^x (x− 1)!

= λe^−λ X∞ x=1

λ^x−1 (x− 1)!

= λe^−λ X∞ x=0

λ^x x! = λ.

E(N²) = X∞ x=0

x²f_N(x) = e^−λ X∞ x=0

x²λ^x x!

= e^−λ X∞ x=1

xλ^x

(x− 1)! (again, the x = 0 contributes nothing)

= λe^−λ X∞ x=1

xλ^x−1 (x− 1)!

= λe^−λ X∞ x=0

(x + 1)λ^x x!

= λe^−λ X∞ x=0

xλ^x

x! + λe^−λ X∞ x=0

λ^x x!

= λE(N ) + λ = λ² + λ.

6. The number of calls received during the night by a doctor on call is a random variable having Poisson distribution with mean 2.5. Suppose she is on call on Friday, Saturday and Sunday nights, and that the numbers of calls on each of these nights are independent Poisson random variables with mean 2.5.

(a) What is the probability that all 3 nights are undisturbed? (You should be able to think of 2 ways to do this!)

(b) What is the probability that she receives no more than 3 calls in total?

(c) Use the normal approximation for (b) and compare with the exact answer.

Solution:

If Y denotes the total number of calls, then since Y is the sum of 3 independent Poisson random variables each with mean 2.5, Y is a Poisson random variable with mean 7.5.

(a) P (Y = 0) = e^−7.5= 0.00055. [Alternatively, we can consider each night separately, and

P (Y = 0) = P (X₁ = 0, X₂ = 0, X₃ = 0)

= P (X₁ = 0)P (X₂ = 0)P (X₃ = 0) (independence)

= (e^−2.5)³ = e^−7.5,

where X_i is of course the number of calls received on the i^th night.]

(b) P (Y ≤ 3) = 0.0591;

(c) Using the normal approximation in (b), we get P (Y ≤ 3) = P ((Y − 7.5)/√

7.5≤ (3 − 7.5)/√ 7.5)

≈ Φ(−1.64) = 0.0505.