Midterm 2 Review Problems Solutions

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) 𝜋𝜋𝑟𝑟 2 𝑃𝑃(dart lands in region 𝐴𝐴) = Therefore, for any 𝑧𝑧 in 0 ≤ 𝑧𝑧 ≤ 𝑟𝑟, we have and hence 𝑃𝑃(𝑍𝑍 ≤ 𝑧𝑧) = 𝜋𝜋𝑧𝑧 2 𝑧𝑧 2 = 𝜋𝜋𝑟𝑟 2 𝑟𝑟 2 𝟎𝟎 𝒛𝒛𝟐𝟐 𝐹𝐹𝑍𝑍 (𝑧𝑧) = � 𝒓𝒓𝟐𝟐 𝟏𝟏 if 𝒛𝒛 < 𝟎𝟎 if 𝟎𝟎 ≤ 𝒛𝒛 ≤ 𝒓𝒓 𝟐𝟐 𝑓𝑓𝑍𝑍 (𝑧𝑧) = � 𝒓𝒓𝟐𝟐 𝟎𝟎 if 𝟎𝟎 ≤ 𝒛𝒛 ≤ 𝒓𝒓 if 𝒛𝒛 > 𝒓𝒓 ∞ 𝑟𝑟 𝑑𝑑 𝐸𝐸 [𝑍𝑍] = � 𝑧𝑧𝑓𝑓𝑍𝑍 (𝑧𝑧) 𝑑𝑑 = � −∞ 0 𝑟𝑟 2𝑧𝑧 2 2 𝑧𝑧 3 2𝑟𝑟 3 𝟐𝟐 𝑑𝑑 = 2 ⋅ � = 2 = 𝑟𝑟 2 𝑟𝑟 3 0 3𝑟𝑟 𝟑𝟑 ■ 𝑉𝑉 c. Find the variance 𝑉𝑉𝑉𝑉 (𝑍𝑍). Solution: ■ 𝟐𝟐 b. Find the mean 𝐸𝐸[𝑍𝑍]. Solution: otherwise 𝑑𝑑 𝟐𝟐 Differentiating with respect to 𝑧𝑧 gives us 𝑑𝑑 −∞ 𝑉𝑉𝑉𝑉 (𝑍𝑍) = 𝐸𝐸 [𝑍𝑍 2 ] − 𝐸𝐸 [𝑍𝑍]2 = 𝑉𝑉 and hence 𝑟𝑟 0 𝑟𝑟 2𝑧𝑧 3 2 𝑧𝑧 4 𝑟𝑟 2 � 𝑑𝑑 = 2 ⋅ = 𝑟𝑟 2 𝑟𝑟 4 0 2 𝑑𝑑 ∞ 𝐸𝐸 [𝑍𝑍 2 ] = � 𝑧𝑧 2 𝑓𝑓𝑍𝑍 (𝑧𝑧) 𝑑𝑑 = � 𝑟𝑟 2 2𝑟𝑟 2 1 4 𝒓𝒓𝟐𝟐 − � � = � − � 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) � 9 𝑃𝑃(𝑌𝑌 > 77) = 𝑃𝑃 �𝑋𝑋 > = 𝑃𝑃 (𝑋𝑋 > 25) = 𝑃𝑃 � = 𝑃𝑃 � 𝑋𝑋 − 10 25 − 10 > � 10 10 𝑋𝑋 − 10 > 1.5� 10 = 1 − 𝑃𝑃 � 𝑋𝑋 − 10 ≤ 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 𝑃𝑃(𝑌𝑌 > 77) = 𝑃𝑃 � = 𝑃𝑃 � 𝑌𝑌 − 50 77 − 50 > � 18 18 𝑌𝑌 − 50 > 1.5� 18 = 1 − 𝑃𝑃 � 𝑌𝑌 − 50 ≤ 1.5� 18 = 1 − Φ(1.5) = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 ■ 3. Let 𝑋𝑋 and 𝑌𝑌 be jointly continuous random variables, and suppose 1 𝑓𝑓𝑋𝑋 |𝑌𝑌 (𝑥𝑥|𝑦𝑦) = � 𝑦𝑦 0 and 𝑓𝑓𝑌𝑌 (𝑦𝑦) = � 2𝑦𝑦 0 if 0 < 𝑦𝑦 ≤ 1 and 1 − 𝑦𝑦 ≤ 𝑥𝑥 ≤ 1 otherwise if 0 < 𝑦𝑦 ≤ 1 . 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 𝑓𝑓𝑋𝑋 ,𝑌𝑌 (𝑥𝑥, 𝑦𝑦) = � 𝟐𝟐 𝟎𝟎 if 𝟎𝟎 < 𝒚𝒚 ≤ 𝟏𝟏 and 𝟏𝟏 − 𝒚𝒚 ≤ 𝒙𝒙 ≤ 𝟏𝟏 otherwise b. Determine the marginal PDF 𝑓𝑓𝑋𝑋 (𝑥𝑥). Solution: For any 𝑥𝑥 in the range 0 ≤ 𝑥𝑥 ≤ 1 we have ∞ so ■ 1 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = � 𝑓𝑓𝑋𝑋,𝑌𝑌 (𝑥𝑥, 𝑦𝑦) 𝑑𝑑𝑑𝑑 = � 2 𝑑𝑑𝑑𝑑 = 2𝑦𝑦|11−𝑥𝑥 = 2�1 − (1 − 𝑥𝑥 )� = 2𝑥𝑥 𝑓𝑓𝑋𝑋 (𝑥𝑥 ) = � −∞ 𝟐𝟐𝟐𝟐 𝟎𝟎 if 𝟎𝟎 ≤ 𝒙𝒙 ≤ 𝟏𝟏 otherwise c. Determine the expected value 𝐸𝐸[𝑋𝑋]. Solution: ∞ 1 1−𝑥𝑥 𝐸𝐸 [𝑋𝑋] = � 𝑥𝑥𝑓𝑓𝑋𝑋 (𝑥𝑥 ) 𝑑𝑑𝑑𝑑 = � 2𝑥𝑥 2 𝑑𝑑𝑑𝑑 = −∞ 0 ■ 2 3 1 𝟐𝟐 𝑥𝑥 � = 3 𝟑𝟑 0 ■ 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 = ⋅ � 𝑦𝑦 2 1−𝑦𝑦 −∞ 1−𝑦𝑦 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 𝑓𝑓𝑌𝑌 (𝑦𝑦) = with 𝑎𝑎 = 𝑏𝑏 = 1, we have 1 𝑦𝑦 − 𝑏𝑏 𝑓𝑓𝑋𝑋 � � |𝑎𝑎| 𝑎𝑎 𝑓𝑓𝑌𝑌 (𝑦𝑦) = 𝑓𝑓𝑋𝑋 (𝑦𝑦 − 1) =� =� 𝝀𝝀𝒆𝒆−𝝀𝝀(𝒚𝒚−𝟏𝟏) 𝟎𝟎 𝝀𝝀𝒆𝒆𝝀𝝀 ⋅ 𝒆𝒆−𝝀𝝀𝝀𝝀 𝟎𝟎 if 𝒚𝒚 − 𝟏𝟏 ≥ 𝟎𝟎 𝒚𝒚 − 𝟏𝟏 < 𝟎𝟎 if 𝒚𝒚 ≥ 𝟏𝟏 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) 𝟎𝟎 𝝀𝝀𝒆𝒆𝝀𝝀 ⋅ 𝒆𝒆−𝝀𝝀𝝀𝝀 𝟎𝟎 if 𝒚𝒚 − 𝟏𝟏 ≥ 𝟎𝟎 𝒚𝒚 − 𝟏𝟏 < 𝟎𝟎 if 𝒚𝒚 ≥ 𝟏𝟏 𝒚𝒚 < 𝟏𝟏 ■