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UCSD ECE 153 Handout # Prof. Young-Han Kim Wednesday, October 29, 2008
Solutions to Old Midterm Exam
- Light bulbs (20 points). Alice and Bob go shopping for light bulbs. Alice buys two regular light bulbs with life span distributed according to the Exp(1) distribution; she will use these two bulbs one by one. Bob goes for one higher-end bulb with life span distributed according to the Exp(1/2) distribution. We will denote X 1 and X 2 for the life spans of Alice’s bulbs and denote Y for that of Bob’s bulb. We assume that X 1 , X 2 , and Y are independent of each other.
(a) (5 points) What is the pdf of the total life span of Alice’s bulbs, i.e., the pdf of X 1 + X 2? (b) (5 points) Compare the expected life spans of Alice’s choice (two cheap bulbs used sequentially) and Bob’s choice (one expensive bulb). (c) (10 points) What is the probability that Bob’s bulb will outlive Alice’s bulbs?
You may find the following facts useful.
- The pdf of an Exp(λ) random variable X is
fX (x) =
{ λe−λx, x ≥ 0 , 0 , otherwise.
∫ (^) ∞
0
xe−λxdx = 1 λ^2.
Solution:
(a) Let Z = X 1 + X 2. Since X 1 and X 2 are independent,
fZ (z) = (fX 1 ∗ fX 2 )(z) =
∫ (^) z
0
e−xe−(z−x)dx =
∫ (^) z
0
e−z^ dx = ze−z^.
This distribution is called the Gamma distribution. (b) E(X 1 + X 2 ) = E(X 1 ) + E(X 2 ) = 1 + 1 = 2 = E(Y ). (c) This is the probability that Y ≥ X 1 + X 2. There are many ways to calculate P(Y ≥ X 1 + X 2 ).
Method I: As in part (a), let Z = X 1 + X 2. Since Y and Z are independent, the joint distribution of Y and Z is given by
fY,Z (y, z) =
e−^
1 2 y^ · ze−z
for y, z ≥ 0. Therefore,
P(Y ≥ Z) =
{y≥z}
fY,Z (y, z)dydz
0
z
e−^
(^12) y · ze−z^ dydz
0
e−^
(^12) z · ze−z^ dz
0
ze−^
3 2 z^ dz
Method II: We use the law of total probability as follows:
P (Y ≥ X 1 + X 2 ) =
P (Y ≥ X 1 + X 2 |X 1 = x 1 , X 2 = x 2 )fX 1 ,X 2 (x 1 , x 2 )dx 1 dx 2
P (Y ≥ x 1 + x 2 |X 1 = x 1 , X 2 = x 2 )fX 1 ,X 2 (x 1 , x 2 )dx 1 dx 2
P (Y ≥ x 1 + x 2 )fX 1 ,X 2 (x 1 , x 2 )dx 1 dx 2
0
0
e−^
(^12) (x 1 +x 2 ) e−x^1 e−x^2 dx 1 dx 2
0
e−^
(^32) x dx
(a) (5 points) Find the conditional pmf pY |X (y|x) of Y given X. (b) (5 points) Find the joint pmf pX,Y (x, y) of X and Y. (c) (5 points) Find the marginal pmf pY (y) of Y. (d) (5 points) Find the conditional pmf pX|Y (x|y) of X given Y. (e) (5 points) Find the optimal estimator g(Y ) that minimize the mean square error E [(X − g(Y ))^2 ]. (f) (5 points) What is the corresponding mean square error? (g) (5 points) Find the optimal decoder D(Y ) that minimizes the error probability P{X 6 = D(Y )}. (h) (5 points) What is the corresponding probability of error?
Solution:
(a) pY |X (+1|+1) = pY |X (− 1 |−1) = 3/ 4. pY |X (− 1 |+1) = pY |X (+1|−1) = 1/ 4.
(b) pX,Y (+1, +1) = pX,Y (− 1 , −1) = 3/ 8. pX,Y (− 1 , +1) = pX,Y (+1, −1) = 1/ 8.
(c) pY (+1) = pY (−1) = 1/ 2. (d) pX|Y (+1|+1) = pX|Y (− 1 |−1) = 3/ 4. pX|Y (− 1 |+1) = pX|Y (+1|−1) = 1/ 4.
(e) g(+1) = E(X|Y = +1) = +1 × (3/4) + (−1) × (1/4) = +1/ 2. g(−1) = E(X|Y = −1) = − 1 × (3/4) + (+1) × (1/4) = − 1 / 2.
(f)
E(X − g(Y ))^2 =
x,y
(x − g(y))^2 pX,Y (x, y)
= 2 × (3/8) × (1/2)^2 + 2 × (1/8) × (3/2)^2 = 3/ 4.
(g) Since pX|Y (+1|+1) > pX|Y (− 1 |+1) and pX|Y (− 1 |−1) > pX|Y (+1|−1), D(Y ) = Y , i.e., D(+1) = +1 and D(−1) = − 1. (h) P(X 6 = D(Y )) = P(X 6 = Y ) = 1/ 4.
