MATH1715 Tutee Revision Questions

Response: This is very similar to the question asked on 16/12/2010. The answer can be
obtained using a tree diagram or as indicated on 16/10/2010. Thus let Ar denote player
A’s draw on stage r, r = 1, 3, 5, and Bs denote player B’s draw on stage s, s = 2, 4.
P{A wins} = P{A1 = S} + P{(A1 = F) ∩ (B2 = F) ∩ (A3 = S)} + · · ·
where P{A1 = S} = 1
, and 5
and so on.
P{(A1 = F) ∩ (B2 = F) ∩ (A3 = S)} = 4 3 1
× ×
5 4 3 ,
Question: 3/1/2011
A stick of length L is broken into two parts at a random point chosen with uniform distribution.
What is the expected length of the shortest part?
Response: This is quite hard.
Let the stick be broken at a point x so the two halves have length x and L − x for
0 < x < L. Without loss of generality we can suppose that the shortest length is x for
0 < x < 1L.
(If this is the longer length, then use the other piece which has length
2
uniformly distributed between 0 and 1
L.) This shorter piece has a uniform distribution
2
between 0 and 1L
so the probability density function is f(x) = 2/L. (This is the key part
2
to understand!)
The mean length is
E[X] =
� 1
2 L
0
xf(x)dx =
� 1
2 L
0
2x
dx =
L
� x 2
L
�1
2 L
0
= 1
4 L.
Question: 3/1/2011
Could you help me with this questions please about correlation coefficients.
Let X be a random variable with mean 1 and variance 2. Let Y = −2X + 3. What is
the correlation coefficient between X and Y ?
Response: You can see that X and Y are linearly related. The negative slope −2 shows
that as X increases, then Y decreases. The correlation corr(X, Y ) = −1.
More formally
Thus
Var[Y ] = (−2) 2 Var[X] = 8, cov(X, Y ) = cov(X, −2X + 3) = −2Var[X] = −4.
corr(X, Y ) =
Question: 1/1/2011
Please help me with the following question.
cov(X, Y )
� Var[X]Var[Y ] = −4
√ 2 × 8 = −1.
8