University of California, Los Angeles Department of Statistics
Statistics 100B Instructor: Nicolas Christou Practice 1 – solutions
EXERCISE 1
Find the distribution of the random variable X for each of the following moment-generating functions: a. MX(t)=1et+25.Xfollowsbinomialwithn=5,p=1.
333
t 2−et
1 et
2 . X followsgeometricwithp= 1.
2
= e
= e2(et−1). X follows Poisson with λ = 2.
=
b. M (t) X
c. MX(t)
EXERCISE
Let MX (t) =
a. To find E(X) we need the first derivative of MX(t) w.r.t. t:
M′ (t) = 1et + 22e2t + 33e3t. The first moment is the previous derivastive evaluated at t = 0. The result is E(X) = X666
1+4+9=7. 6663
1− 1 et 2
2
1 et + 2 e2t + 3 e3t be the moment-generating function of a random variable X. 666
b. To find var(X) we need the second moment. We need the second derivative of MX(t) w.r.t. t.
M′′(t)= 1et+82e2t+273e3t. Andatt=0weobtainthesecondmoment. ItisequaltoEX2 = 1 +8 +27 =6. The
X666 666 variance is equal to: var(X) = EX2 − (E(X))2 = 6 − ( 7 )2 = 5 .
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c. From the definition of moment-generating functions MX(t) = EetX we see that X is discrete with possible values 1, 2, and 3, and corresponding probabilities 1 , 2 , and 3 .
666
Let X follow the Poisson probability distribution with parameter λ. Its moment-generating function is MX(t) = eλ(et−1).
EXERCISE 3
X−λ
a. The moment-generating function of Z = √ is:
λ
e
t λ√√
−√ t t − λt λ(e λ−1)
tt2t3 √(√)(√)
MZ(t)=MX−λ(t)=e λ MX(√ )⇒MZ(t)=e √λ λ
.
b. Using the series expansion of
t
e√λ=1+ λ+ λ + λ +···
we get:
1! 2! 3!
√ t t2 t3 √ √ t2 λt3
MZ (t) = e− Therefore,
λt−λ+λ[1+ √λ + 2λ + 3!(√λ)3 +···] = e− λt−λ+λ+ λt+ 2 + 3!(√λ)3 +···
In other words, as λ → ∞, the ratio Z = √ EXERCISE 4
X−λ
λ
1 t2 lim MZ(t)=e2 .
λ→∞
converges to the standard normal distribution.
Here we use the normal approximation to Poisson. It is given that X follows the Poisson distribution with λ = 100. We know
X−λ
that for large λ the ratio Z = √ follows the standard normal distribution. Therefore:
λ
110.5 − 100
P(X ≤110)=P(Z < √ The exact probability is
110
e−100100x
P(X ≤ 110) =
The approximation is not bad!
EXERCISE 5
= ··· = 0.8529.
x! x=0
100
)=P(Z <1.05)=0.8531.
μt+ 1 σ2t2
We know that the moment generating function of a normal random variable is e 2 . Because the sample is i.i.d the
moment generating function of T = ni=1 Xi is:
T t (X1+···+Xn)t X1t Xnt nμt+ 1 nσ2t2
MT(t) = Ee = Ee = Ee ···Ee ⇒ MT(t) = e 2 . This is the moment generating function of a normal random variable with mean nμ and variance nσ2. Therefore T ∼ N(nμ,σ√n).
EXERCISE 6
The two sample are independent with X ∼ N(μ1,σ1) and Y ∼ N(μ2,σ2). a. E(X ̄−Y ̄)=E(X ̄)−E(Y ̄)=μ1−μ2.
̄ ̄ ̄ ̄ ̄ ̄σ12σ2
b. Var(X − Y) = Var(X) + Var(Y) − 2Cov(X,Y) = n + m . The covariance is zero because the two samples are
independent.
c. M ̄ ̄ =Ee(X ̄−Y ̄)t =EeX ̄tEe−Y ̄t =M (t )···M (t )M (−t)···M (−t)=
X−Y
m ( μ 1 t + 1 σ 2 t 2 ) − n ( μ 2 t − 1 σ 2 t 2 )
X1 m Xn m Y1 n Yn n 1 2 σ 12 σ 2 2
e m 2 1 m2 e n 2 2 n2
= et(μ1−μ2)+2t ( m + n ). This is the moment generating function of a normal σ12 σ2
random variable with mean μ1 − μ2 and variance m + n .
d. Itisgiventhatn=mandσ12=2,σ2=2.5.WewantP(−1