CSE 107 Midterm 2 Review Problems PDF

CSE 107 Midterm 2 Review Problems Solutions 1. Bob throws a dart at a circular target of radius 𝑟𝑟. He hits the target with certainty, but is equally likely to hit any point within the target. Let 𝑍𝑍 be the distance from Bob’s dart to the center of the target. a. Find the CDF 𝐹𝐹𝑍𝑍 (𝑧𝑧) and t...

CSE 107 Midterm 2 Review Problems Solutions 1. Bob throws a dart at a circular target of radius 𝑟𝑟. He hits the target with certainty, but is equally likely to hit any point within the target. Let 𝑍𝑍 be the distance from Bob’s dart to the center of the target. a. Find the CDF 𝐹𝐹𝑍𝑍 (𝑧𝑧) and the PDF 𝑓𝑓𝑍𝑍 (𝑧𝑧). Solution: Since Bob is equally likely to hit any point, the probability of the dart landing within any region of the target is proportional to the area of that region. Since the area of the target is 𝜋𝜋𝑟𝑟 2 we have area(𝐴𝐴) 𝑃𝑃(dart lands in region 𝐴𝐴) = 𝜋𝜋𝑟𝑟 2 Therefore, for any 𝑧𝑧 in 0 ≤ 𝑧𝑧 ≤ 𝑟𝑟, we have 𝜋𝜋𝑧𝑧 2 𝑧𝑧 2 𝑃𝑃(𝑍𝑍 ≤ 𝑧𝑧) = = 𝜋𝜋𝑟𝑟 2 𝑟𝑟 2 and hence 𝟎𝟎 if 𝒛𝒛 < 𝟎𝟎 𝒛𝒛𝟐𝟐 𝐹𝐹𝑍𝑍 (𝑧𝑧) = if 𝟎𝟎 ≤ 𝒛𝒛 ≤ 𝒓𝒓 𝒓𝒓𝟐𝟐 𝟏𝟏 if 𝒛𝒛 > 𝒓𝒓 Differentiating with respect to 𝑧𝑧 gives us 𝟐𝟐𝟐𝟐 𝑓𝑓𝑍𝑍 (𝑧𝑧) = 𝒓𝒓𝟐𝟐 if 𝟎𝟎 ≤ 𝒛𝒛 ≤ 𝒓𝒓 𝟎𝟎 otherwise b. Find the mean 𝐸𝐸[𝑍𝑍]. Solution: ∞ 𝑟𝑟 𝑟𝑟 2𝑧𝑧 2 2 𝑧𝑧 3 2𝑟𝑟 3 𝟐𝟐𝟐𝟐 𝐸𝐸 [𝑍𝑍] = 𝑧𝑧𝑓𝑓𝑍𝑍 (𝑧𝑧) 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = 2 ⋅ = 2 = −∞ 0 𝑟𝑟 2 𝑟𝑟 3 0 3𝑟𝑟 𝟑𝟑 c. Find the variance 𝑉𝑉𝑉𝑉𝑉𝑉(𝑍𝑍). Solution: ∞ 𝑟𝑟 𝑟𝑟 2𝑧𝑧 3 2 𝑧𝑧 4 𝑟𝑟 2 𝐸𝐸 [𝑍𝑍 2 ] = 𝑧𝑧 2 𝑓𝑓𝑍𝑍 (𝑧𝑧) 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = 2 ⋅ = −∞ 0 𝑟𝑟 2 𝑟𝑟 4 0 2 and hence 𝑟𝑟 2 2𝑟𝑟 2 1 4 𝒓𝒓𝟐𝟐 𝑉𝑉𝑉𝑉𝑉𝑉(𝑍𝑍) = 𝐸𝐸 [𝑍𝑍 2 ] − 𝐸𝐸 [𝑍𝑍]2 = − = − 𝑟𝑟 2 = 2 3 2 9 𝟏𝟏𝟏𝟏 2. A city’s temperature in degrees Celsius is modeled as a normal random variable 𝑋𝑋 with mean 10 and standard deviation 10. Let 𝑌𝑌 be its temperature in Fahrenheit, where 𝑋𝑋 and 𝑌𝑌 are related by 5(𝑌𝑌 − 32) 𝑋𝑋 =. 9 What is the probability that the temperature is above 77 degrees Fahrenheit? Solution: 5(77 − 32) 𝑃𝑃(𝑌𝑌 > 77) = 𝑃𝑃 𝑋𝑋 > 9 = 𝑃𝑃 (𝑋𝑋 > 25) 𝑋𝑋 − 10 25 − 10 = 𝑃𝑃 > 10 10 𝑋𝑋 − 10 = 𝑃𝑃 > 1.5 10 𝑋𝑋 − 10 = 1 − 𝑃𝑃 ≤ 1.5 10 = 1 − Φ(1.5) = 1 − 0.9332 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 Alternate Solution: We have 𝑌𝑌 = (9𝑋𝑋 + 160)/5, and therefore 𝑌𝑌 is normal with mean (9 ⋅ 10 + 160)/5 = 50, and standard deviation (9/5) ⋅ 10 = 18. Thus 𝑌𝑌 − 50 77 − 50 𝑃𝑃(𝑌𝑌 > 77) = 𝑃𝑃 > 18 18 𝑌𝑌 − 50 = 𝑃𝑃 > 1.5 18 𝑌𝑌 − 50 = 1 − 𝑃𝑃 ≤ 1.5 18 = 1 − Φ(1.5) = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 3. Let 𝑋𝑋 and 𝑌𝑌 be jointly continuous random variables, and suppose 1 if 0 < 𝑦𝑦 ≤ 1 and 1 − 𝑦𝑦 ≤ 𝑥𝑥 ≤ 1 𝑓𝑓𝑋𝑋 |𝑌𝑌 (𝑥𝑥|𝑦𝑦) = 𝑦𝑦 0 otherwise and 2𝑦𝑦 if 0 < 𝑦𝑦 ≤ 1 𝑓𝑓𝑌𝑌 (𝑦𝑦) =. 0 otherwise Hint: Before you do the following problems, draw a picture of the region defined by the inequalities 0 < 𝑦𝑦 ≤ 1 and 1 − 𝑦𝑦 ≤ 𝑥𝑥 ≤ 1. a. Determine the joint PDF 𝑓𝑓𝑋𝑋 ,𝑌𝑌 (𝑥𝑥, 𝑦𝑦). Solution: The region defined by the inequalities 0 < 𝑦𝑦 ≤ 1 and 1 − 𝑦𝑦 ≤ 𝑥𝑥 ≤ 1 is the triangle with vertices (0, 1), (1, 0) and (1, 1), which has area 1/2. Therefore 𝟐𝟐 if 𝟎𝟎 < 𝒚𝒚 ≤ 𝟏𝟏 and 𝟏𝟏 − 𝒚𝒚 ≤ 𝒙𝒙 ≤ 𝟏𝟏 𝑓𝑓𝑋𝑋 ,𝑌𝑌 (𝑥𝑥, 𝑦𝑦) = 𝟎𝟎 otherwise b. Determine the marginal PDF 𝑓𝑓𝑋𝑋 (𝑥𝑥). Solution: For any 𝑥𝑥 in the range 0 ≤ 𝑥𝑥 ≤ 1 we have ∞ 1 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = 𝑓𝑓𝑋𝑋,𝑌𝑌 (𝑥𝑥, 𝑦𝑦) 𝑑𝑑𝑑𝑑 = 2 𝑑𝑑𝑑𝑑 = 2𝑦𝑦|11−𝑥𝑥 = 2 1 − (1 − 𝑥𝑥 ) = 2𝑥𝑥 −∞ 1−𝑥𝑥 so 𝟐𝟐𝟐𝟐 if 𝟎𝟎 ≤ 𝒙𝒙 ≤ 𝟏𝟏 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = 𝟎𝟎 otherwise c. Determine the expected value 𝐸𝐸[𝑋𝑋]. Solution: ∞ 1 2 3 1 𝟐𝟐 𝐸𝐸 [𝑋𝑋] = 𝑥𝑥𝑓𝑓𝑋𝑋 (𝑥𝑥 ) 𝑑𝑑𝑑𝑑 = 2𝑥𝑥 2 𝑑𝑑𝑑𝑑 = 𝑥𝑥 = −∞ 0 3 0 𝟑𝟑 d. Determine the conditional expectation 𝐸𝐸[𝑋𝑋|𝑌𝑌 = 𝑦𝑦]. Solution: ∞ 1 1 𝑥𝑥 1 𝑥𝑥 2 𝐸𝐸 [𝑋𝑋|𝑌𝑌 = 𝑦𝑦] = 𝑥𝑥𝑓𝑓𝑋𝑋 |𝑌𝑌 (𝑥𝑥|𝑦𝑦) 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = ⋅ −∞ 1−𝑦𝑦 𝑦𝑦 𝑦𝑦 2 1−𝑦𝑦 𝟏𝟏 − (𝟏𝟏 − 𝒚𝒚)𝟐𝟐 𝟐𝟐𝟐𝟐 − 𝒚𝒚𝟐𝟐 𝟐𝟐 − 𝒚𝒚 𝒚𝒚 = = = = 𝟏𝟏 − 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐 𝟐𝟐 4. Let 𝑋𝑋 be an exponential random variable with parameter 𝜆𝜆, and let 𝑌𝑌 = 𝑋𝑋 +1. Determine the PDF 𝑓𝑓𝑌𝑌 (𝑦𝑦). Solution: Using the formula for the PDF of 𝑌𝑌 = 𝑎𝑎𝑎𝑎 + 𝑏𝑏 derived in class 1 𝑦𝑦 − 𝑏𝑏 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑓𝑓𝑋𝑋 |𝑎𝑎| 𝑎𝑎 with 𝑎𝑎 = 𝑏𝑏 = 1, we have 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑓𝑓𝑋𝑋 (𝑦𝑦 − 1) 𝝀𝝀𝒆𝒆−𝝀𝝀(𝒚𝒚−𝟏𝟏) if 𝒚𝒚 − 𝟏𝟏 ≥ 𝟎𝟎 = 𝟎𝟎 𝒚𝒚 − 𝟏𝟏 < 𝟎𝟎 𝝀𝝀𝒆𝒆𝝀𝝀 ⋅ 𝒆𝒆−𝝀𝝀𝝀𝝀 if 𝒚𝒚 ≥ 𝟏𝟏 = 𝟎𝟎 𝒚𝒚 < 𝟏𝟏 Alternate Solution: Using the two-step process, we have 𝐹𝐹𝑌𝑌 (𝑦𝑦) = 𝑃𝑃(𝑌𝑌 ≤ 𝑦𝑦) = 𝑃𝑃(𝑋𝑋 + 1 ≤ 𝑦𝑦) = 𝑃𝑃 (𝑋𝑋 ≤ 𝑦𝑦 − 1) = 𝐹𝐹𝑋𝑋 (𝑦𝑦 − 1) Differentiating with respect to 𝑦𝑦 gives 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑓𝑓𝑋𝑋 (𝑦𝑦 − 1) 𝝀𝝀𝒆𝒆−𝝀𝝀(𝒚𝒚−𝟏𝟏) if 𝒚𝒚 − 𝟏𝟏 ≥ 𝟎𝟎 = 𝟎𝟎 𝒚𝒚 − 𝟏𝟏 < 𝟎𝟎 𝝀𝝀𝒆𝒆𝝀𝝀 ⋅ 𝒆𝒆−𝝀𝝀𝝀𝝀 if 𝒚𝒚 ≥ 𝟏𝟏 = 𝟎𝟎 𝒚𝒚 < 𝟏𝟏 5. Alice is at the casino again, with a choice of two games. The first returns winnings (positive or negative) that are normally distributed with parameters 𝜇𝜇 = 1 and 𝜎𝜎 = 2. The second is uniformly distributed with winnings in the range −1 to 2. (All amounts are in dollars.) She flips a coin with 𝑃𝑃(head) = 𝑝𝑝 to decide which game to play. If heads, she plays the first game, and if tails, she plays the second. Determine her expected winnings, in terms of 𝑝𝑝. Solution: Let 𝑋𝑋 be the amount of money Alice wins, and let 𝐴𝐴 be the event that the coin flip is heads. We are given that 𝑃𝑃(𝐴𝐴) = 𝑝𝑝 and 𝑃𝑃(𝐴𝐴𝑐𝑐 ) = 1 − 𝑝𝑝. We also know that 𝐸𝐸 [𝑋𝑋|𝐴𝐴] = 1 and 𝐸𝐸 [𝑋𝑋|𝐴𝐴𝑐𝑐 ] = 1/2. The total expectation theorem now gives 𝐸𝐸 [𝑋𝑋] = 𝑝𝑝𝑝𝑝 [𝑋𝑋|𝐴𝐴] + (1 − 𝑝𝑝)𝐸𝐸[𝑋𝑋|𝐴𝐴𝑐𝑐 ] 1 = 𝑝𝑝 ⋅ 1 + (1 − 𝑝𝑝) ⋅ 2 𝒑𝒑 + 𝟏𝟏 = 𝟐𝟐 6. Let 𝑌𝑌 be a normal random variable with variance 1, and with mean another random variable 𝑋𝑋. Suppose 𝑋𝑋 is continuous uniform on the interval [1, 3]. a. Find the PDF 𝑓𝑓𝑌𝑌 (𝑦𝑦). Solution: We are given that 1/2 if 1 ≤ 𝑥𝑥 ≤ 3 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = 0 otherwise and 1 2/2 𝑒𝑒 −(𝑦𝑦−𝑥𝑥) if 1 ≤ 𝑥𝑥 ≤ 3 and 𝑦𝑦 ∈ ℝ 𝑓𝑓𝑌𝑌|𝑋𝑋 (𝑦𝑦|𝑥𝑥 ) = √2𝜋𝜋 0 otherwise Therefore ∞ ∞ 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑓𝑓𝑋𝑋,𝑌𝑌 (𝑥𝑥, 𝑦𝑦) 𝑑𝑑𝑑𝑑 = 𝑓𝑓𝑌𝑌|𝑋𝑋 (𝑦𝑦|𝑥𝑥) ⋅ 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) 𝑑𝑑𝑑𝑑 −∞ −∞ 3 1 𝑒𝑒 −(𝑥𝑥−𝑦𝑦) 2/2 𝑢𝑢 = 𝑥𝑥 − 𝑦𝑦, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 substitute: 𝑥𝑥 = 3 ⇒ 𝑢𝑢 = 3 − 𝑦𝑦 2 √2𝜋𝜋 1 𝑥𝑥 = 1 ⇒ 𝑢𝑢 = 1 − 𝑦𝑦 3−𝑦𝑦 2 1 𝑒𝑒 −𝑢𝑢 /2 = 𝑑𝑑𝑑𝑑 2 √2𝜋𝜋 1−𝑦𝑦 𝟏𝟏 = 𝚽𝚽(𝟑𝟑 − 𝒚𝒚) − 𝚽𝚽(𝟏𝟏 − 𝒚𝒚) 𝟐𝟐 b. Find the conditional PDF 𝑓𝑓𝑋𝑋|𝑌𝑌 (𝑥𝑥|𝑦𝑦). Solution: Baye’s rule gives us 𝑓𝑓𝑌𝑌|𝑋𝑋 (𝑦𝑦|𝑥𝑥) ⋅ 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) 𝑓𝑓𝑋𝑋 |𝑌𝑌 (𝑥𝑥|𝑦𝑦) = 𝑓𝑓𝑌𝑌 (𝑦𝑦) 1 −(𝑦𝑦−𝑥𝑥)2/2 ⎧ 𝑒𝑒 ⎪ √2𝜋𝜋 if 1 ≤ 𝑥𝑥 ≤ 3 and 𝑦𝑦 ∈ ℝ = Φ(3 − 𝑦𝑦) − Φ(1 − 𝑦𝑦) ⎨ ⎪ ⎩0 otherwise c. Suppose we sample 𝑌𝑌 and get 𝑌𝑌 = 3. What is the probability that 𝑋𝑋 ≤ 2 ? Solution: 2 𝑃𝑃(𝑋𝑋 ≤ 2 | 𝑌𝑌 = 3) = 𝑓𝑓𝑋𝑋|𝑌𝑌 (𝑥𝑥|3) 𝑑𝑑𝑑𝑑 −∞ 2 𝑢𝑢 = 𝑥𝑥 − 3 2 1/√2𝜋𝜋 𝑒𝑒 −(𝑥𝑥−3) /2 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 substitute: 1 Φ(0) − Φ(−2) 𝑥𝑥 = 2 ⇒ 𝑢𝑢 = −1 𝑥𝑥 = 1 ⇒ 𝑢𝑢 = −2 −1 2 ∫−2 1/√2𝜋𝜋 𝑒𝑒 −𝑢𝑢 /2 𝑑𝑑𝑑𝑑 = Φ (0) − Φ(−2) Φ(−1) − Φ(−2) = = 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 Φ(0) − Φ(−2) d. Find 𝐸𝐸[𝑌𝑌]. Solution: ∞ 𝐸𝐸 [𝑌𝑌] = 𝑦𝑦 ⋅ 𝑓𝑓𝑌𝑌 (𝑦𝑦) 𝑑𝑑𝑑𝑑 −∞ 3 2/2 ∞ 1 𝑒𝑒 −(𝑥𝑥−𝑦𝑦) = 𝑦𝑦 ⋅ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 by part (𝑎𝑎) −∞ 2 √2𝜋𝜋 1 3 ∞ 2/2 1 𝑒𝑒 −(𝑦𝑦−𝑥𝑥) = 𝑦𝑦 ⋅ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 the inner integral is 𝑥𝑥 by inspection 2 √2𝜋𝜋 1 −∞ 3 3 1 1 𝑥𝑥 2 1 = 𝑥𝑥 𝑑𝑑𝑑𝑑 = ⋅ = (9 − 1) = 𝟐𝟐 2 2 2 1 4 1 7. Let 𝑋𝑋 and 𝑌𝑌 be independent, jointly continuous random variables, where 𝑋𝑋 is uniform on [𝑎𝑎, 𝑏𝑏] and 𝑌𝑌 is uniform on [𝑐𝑐, 𝑑𝑑]. a. Write the PDFs 𝑓𝑓𝑋𝑋 (𝑥𝑥) and 𝑓𝑓𝑌𝑌 (𝑦𝑦). Solution: 1 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = 𝑏𝑏 − 𝑎𝑎 if 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏 0 otherwise and 1 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑑𝑑 − 𝑐𝑐 if 𝑐𝑐 ≤ 𝑦𝑦 ≤ 𝑑𝑑 0 otherwise b. Write the CDFs 𝐹𝐹𝑋𝑋 (𝑥𝑥) and 𝐹𝐹𝑌𝑌 (𝑦𝑦). Solution: 0 if 𝑥𝑥 < 𝑎𝑎 𝑥𝑥 − 𝑎𝑎 𝐹𝐹𝑋𝑋 (𝑥𝑥) = if 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏 𝑏𝑏 − 𝑎𝑎 1 if 𝑥𝑥 > 𝑏𝑏 and 0 if 𝑦𝑦 < 𝑐𝑐 𝑦𝑦 − 𝑐𝑐 𝐹𝐹𝑌𝑌 (𝑦𝑦) = if 𝑐𝑐 ≤ 𝑦𝑦 ≤ 𝑑𝑑 𝑑𝑑 − 𝑐𝑐 1 if 𝑦𝑦 > 𝑑𝑑 c. Let 𝑍𝑍 = max(𝑋𝑋, 𝑌𝑌). Determine the PDF 𝑓𝑓𝑍𝑍 (𝑧𝑧). (Assume 𝑎𝑎 ≤ 𝑑𝑑 and 𝑐𝑐 ≤ 𝑏𝑏, so that the two intervals overlap.) Solution: Using the independence of 𝑋𝑋 and 𝑌𝑌, 𝐹𝐹𝑍𝑍 (𝑧𝑧) = 𝑃𝑃(𝑍𝑍 ≤ 𝑧𝑧) = 𝑃𝑃(max(𝑋𝑋, 𝑌𝑌) ≤ 𝑧𝑧) = 𝑃𝑃(𝑋𝑋 ≤ 𝑧𝑧, 𝑌𝑌 ≤ 𝑧𝑧) = 𝑃𝑃(𝑋𝑋 ≤ 𝑧𝑧) ⋅ 𝑃𝑃(𝑌𝑌 ≤ 𝑧𝑧) = 𝐹𝐹𝑋𝑋 (𝑧𝑧) ⋅ 𝐹𝐹𝑌𝑌 (𝑧𝑧). From part (b) we have the following two cases. Case 1: 𝑏𝑏 < 𝑑𝑑 0 𝑧𝑧 < max(𝑎𝑎, 𝑐𝑐 ) ⎧ ⎪ (𝑧𝑧 − 𝑎𝑎)(𝑧𝑧 − 𝑐𝑐) max(𝑎𝑎, 𝑐𝑐 ) ≤ 𝑧𝑧 < 𝑏𝑏 𝐹𝐹𝑍𝑍 (𝑧𝑧) = (𝑏𝑏 − 𝑎𝑎)(𝑑𝑑 − 𝑐𝑐) ⎨ 𝑧𝑧 − 𝑐𝑐 ⎪ 𝑑𝑑 − 𝑐𝑐 𝑏𝑏 ≤ 𝑧𝑧 ≤ 𝑑𝑑 ⎩1 𝑧𝑧 > 𝑑𝑑 Case 2: 𝑑𝑑 ≤ 𝑏𝑏 0 𝑧𝑧 < max(𝑎𝑎, 𝑐𝑐 ) ⎧ ⎪ (𝑧𝑧 − 𝑎𝑎)(𝑧𝑧 − 𝑐𝑐) max(𝑎𝑎, 𝑐𝑐 ) ≤ 𝑧𝑧 < 𝑑𝑑 𝐹𝐹𝑍𝑍 (𝑧𝑧) = (𝑏𝑏 − 𝑎𝑎)(𝑑𝑑 − 𝑐𝑐) ⎨ 𝑧𝑧 − 𝑎𝑎 ⎪ 𝑏𝑏 − 𝑎𝑎 𝑑𝑑 ≤ 𝑧𝑧 ≤ 𝑏𝑏 ⎩1 𝑧𝑧 > 𝑏𝑏  Differentiate with respect to 𝑧𝑧 to obtain Case 1: 𝑏𝑏 < 𝑑𝑑 𝟐𝟐𝟐𝟐 − (𝒂𝒂 + 𝒄𝒄) ⎧ 𝐦𝐦𝐦𝐦𝐦𝐦(𝒂𝒂, 𝒄𝒄) ≤ 𝒛𝒛 < 𝒃𝒃 ⎪ (𝒃𝒃 − 𝒂𝒂)(𝒅𝒅 − 𝒄𝒄) 𝒇𝒇𝒁𝒁 (𝒛𝒛) = 𝟏𝟏 ⎨ 𝒃𝒃 ≤ 𝒛𝒛 ≤ 𝒅𝒅 ⎪ 𝒅𝒅 − 𝒄𝒄 ⎩ 𝟎𝟎 otherwise Case 2: 𝑑𝑑 ≤ 𝑏𝑏 𝟐𝟐𝟐𝟐 − (𝒂𝒂 + 𝒄𝒄) ⎧ 𝐦𝐦𝐦𝐦𝐦𝐦(𝒂𝒂, 𝒄𝒄) ≤ 𝒛𝒛 < 𝒅𝒅 ⎪ (𝒃𝒃 − 𝒂𝒂)(𝒅𝒅 − 𝒄𝒄) 𝒇𝒇𝒁𝒁 (𝒛𝒛) = 𝟏𝟏 ⎨ 𝒅𝒅 ≤ 𝒛𝒛 ≤ 𝒃𝒃 ⎪ 𝒃𝒃 − 𝒂𝒂 ⎩ 𝟎𝟎 otherwise d. Let 𝑍𝑍 = 𝑋𝑋 + 𝑌𝑌. Determe the PDF 𝑓𝑓𝑍𝑍 (𝑧𝑧). Solution: By a formula derived in class, we have ∞ 𝑓𝑓𝑍𝑍 (𝑧𝑧) = (𝑓𝑓𝑋𝑋 ∗ 𝑓𝑓𝑌𝑌 )(𝑧𝑧) = 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) ⋅ 𝑓𝑓𝑌𝑌 (𝑧𝑧 − 𝑥𝑥 ) 𝑑𝑑𝑑𝑑 −∞ The integrand is 1/(𝑏𝑏 − 𝑎𝑎)(𝑑𝑑 − 𝑐𝑐 ) and is non-zero if and only if both 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏 and 𝑐𝑐 ≤ 𝑧𝑧 − 𝑥𝑥 ≤ 𝑑𝑑 i.e. 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏 and 𝑧𝑧 − 𝑑𝑑 ≤ 𝑥𝑥 ≤ 𝑧𝑧 − 𝑐𝑐 Hence min(𝑏𝑏,𝑧𝑧−𝑐𝑐) 1 min(𝑏𝑏, 𝑧𝑧 − 𝑐𝑐 ) − max(𝑎𝑎, 𝑧𝑧 − 𝑑𝑑 ) 𝑓𝑓𝑍𝑍 (𝑧𝑧) = 𝑑𝑑𝑑𝑑 = (𝑏𝑏 − 𝑎𝑎)(𝑑𝑑 − 𝑐𝑐 ) (𝑏𝑏 − 𝑎𝑎)(𝑑𝑑 − 𝑐𝑐 ) max(𝑎𝑎,𝑧𝑧−𝑑𝑑) The interval of integration is empty if and only if 𝑏𝑏 < 𝑧𝑧 − 𝑑𝑑 or 𝑧𝑧 − 𝑐𝑐 < 𝑎𝑎. Therefore, it is non- empty if and only if both 𝑎𝑎 ≤ 𝑧𝑧 − 𝑐𝑐 and 𝑧𝑧 − 𝑑𝑑 ≤ 𝑏𝑏, which is equivalent to 𝑎𝑎 + 𝑐𝑐 ≤ 𝑧𝑧 ≤ 𝑏𝑏 + 𝑑𝑑. Thus 𝐦𝐦𝐦𝐦𝐦𝐦(𝒃𝒃, 𝒛𝒛 − 𝒄𝒄) − 𝐦𝐦𝐦𝐦𝐦𝐦(𝒂𝒂, 𝒛𝒛 − 𝒅𝒅) 𝒇𝒇𝒁𝒁 (𝒛𝒛) = if 𝒂𝒂 + 𝒄𝒄 ≤ 𝒛𝒛 ≤ 𝒃𝒃 + 𝒅𝒅 (𝒃𝒃 − 𝒂𝒂)(𝒅𝒅 − 𝒄𝒄) 𝟎𝟎 otherwise