F71SM STATISTICAL METHODS
Tutorial on Section 3 RANDOM VARIABLES: SOLUTIONS
1. μ=E[X]=1×0.5+2×0.25=1
E[X2]=12 ×0.5+22 ×0.25=1.5⇒s2 =Var[X]=1.5−12 =0.5 Pgf: G(t) = 0.25 + 0.5t + 0.25t2 = (1 + t)2/4
∞ 2x 2. μ=E[X]=e−2 x
E[X(X−1)]=e−2 x(x−1) x=0 x!
⇒ E[X2] = 4 + 2 = 6
⇒s2 =Var[X]=6−22 =2
x=0 x! x=0 x! Mgf: M (t) = G et = exp 2 et − 1
∞ 2x−1 =2e−2
∞2y
=2e−2 =2e−2e2 =2
x=0 x!
∞ 2x
y=0 y!
x=1 (x−1)!
∞ 2x−2 =e−222
∞2y =4e−2
=4e−2e2 =4
∞ 2x ∞(2t)x Pgf:G(t)=E tX =txe−2 =e−2
y=0 y! =e−2e2t=e2(t−1)
x=2 (x − 2)!
3. The number of trials required will equal r provided trials 1, 2, 3, . . . , r − 1 consist of 1 S and (r−2) F’s, with trial r being S. Hence P (X = r) = r−1pqr−2×p = (r−1)p2qr−2, r = 2, 3, . . .,
1
where q = 1−p, and so
G(t)=p2(t2 +2qt3 +3q2t4 +···)=p2t2(1−qt)−2 for−1/q
1111
With the given pdf fX(x), we have fY (y) = 2y−1/2 √2πe−y/2 + √2πe−y/2 = √2π y−1/2e−y/2,
for y > 0.
dy2 2
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