This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Econ 319 Spring 2008 TA: Seung Han Yoo Solution to Problem Set 2 1. (#2.31) a. Take a new parameter & = 2 = 6 = 1 = 3 . Then, f ( k ) = (1 = 3) k k ! e & 1 = 3 . Hence, P 1 k =1 f ( k ) = 1 & f (0) = 0 : 28 . b. f (0) = e & & ¡ : 5 . & ¢ & log (0 : 5) . Then, & = 0 : 693 and : 693 £ 30 = 20 : 79 min. 2. (#2.39) a. Since lim x !&1 tan & 1 ( x ) = & ¡ 2 and lim x ! + 1 tan & 1 ( x ) = ¡ 2 , we have lim x !&1 F ( x ) = and lim x ! + 1 F ( x ) = 1 . Note that F ( x ) = 1 ¡ (1+ x 2 ) > , which implies that F ( x ) is increasing and continuous. b. f ( x ) = F ( x ) = 1 ¡ (1+ x 2 ) . c. P ( X > x ) = 0 : 1 . Then, P ( X < x ) = 0 : 9 and : 5 + 1 ¡ tan & 1 ( x ) = 0 : 9 . Hence, x = tan (0 : 4 ¡ ) = 3 : 08 . 3. (#2.40) a. c R 1 x 2 dx = 1 and since R 1 x 2 dx = 1 3 , c = 3 . b. F ( x ) = 3 R x u 2 du = x 3 . c. F (0 : 5) & F (0 : 1) = 0 : 124 . 4. (#2.44) 1 Givne f T ( T ) = &e & &T , f ( k ) = R k +1 k f T ( T ) dT = R k +1 k &e & &T dT = & e & & ( k +1) + e & &k = e & &k & 1 & e & &...

View
Full Document

- Summer '07
- HONG
- Inch