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Lecture

# Tutorial2_soln.pdf

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University of Waterloo

Statistics

STAT 330

Christine Dupont

Fall

Description

Stat 330 - Tutorial 2
In this tutorial, we will use the Binomial series quite often. That is,
n
n X n x n−x
(a + b) = a b .
x=0 x
1. Sol:
(a) By deﬁnition,
X n
E(X (k) = x(x − 1)··· (x − k + 1)p (1 − p)x
x
x=0
(Note that x(x − 1)··· (x − k + 1) = 0 for x = 0,... ,k − 1.)
n
X n! x n−x
= x(x − 1)··· (x − k + 1) p (1 − p) .
x=k x!(n − x)!
We simplify the above form by noting that
n! = n(n − 1)··· (n − k + 1)(n − k)!, x! = x(x − 1)··· (x − k + 1)(x − k)!.
Therefore
n
(k) X n(n − 1)··· (n − k + 1)(n −xk)! n−x
E(X ) = p (1 − p) .
x=k (x − k)!(n − k)!
Changing the index x − k = t, then t = 0,... ,n − k and
X−k X−k
(k) n − k k+t n−k−t (k) k n − k t n−k−t
E(X ) = n(n−1)···(n−k+1) t p (1−p) = n p t p (1−p) .
t=0 t=0
Using the Binomial series, we have
E(X (k)) = nkp .
From this result, we get
2
µ = E(X) = np, E{X(X − 1)} = n(n − 1)p .
Therefore
V ar(X) = E(X )−µ = E{X(X−1)}+E(X)−(np) = n(n−1)p +np−(np) = np(1−p).
(b) By deﬁnition,
n n
tX X txn x n−x X n t x n−x
M(t) = E(e ) = e p (1 − p) = (pe ) (1 − p) .
x=0 x x=0 x
Using Binomial series, we get
M(t) = (pe + 1 − p) with |t| < ∞.
1 Note that
′ t n−1 t ′′ t n−1 t t n−2 t 2
M (t) = n(pe +1−p) pe and M (t) = n(pe +1−p) pe +n(n−1)(pe +1−p) (pe ) .
Therefore
E(X) = M (0) = np, E(X ) = M (0) = np + n(n − 1)p = (np) + np(1 − p),
which imply that V ar(X) = (np) + np(1 − p) − (np) = np(1 − p).
2 2
2. Sol: Let 1 = θ ,2p = 2θ(1 − θ), and3p = (1 − θ) . Note th1t p 2 p 3 p = 1.
(a) The joint pf of X and Y is given
n! x y n−x−y
f(x,y) = P(X = x,Y = y) = p1p2 3 ,
x!y!(n − x − y)!
where x = 0,... ,n, y = 0,... ,n, and x + y ≤ n.
(b) For x = 0,... ,n, the mar

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