HW8_solutions - HW#8 Solutions AMS 311 Probability Theory...

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HW#8 Solutions AMS 311 Probability Theory (Spring 2011) Part 1 (10 points): PRINT your name and StonyBrook ID ( Last, First, #ID ) on the top on your work sheet. Staple your HW if more than 1 pages. Part 2: Problems (90 points) Write down your work step by step to get full score. You do not need to evaluate arithmetic expressions or integrals, if they are fully specified. For example, you may leave in this form. (1) (24 points) (a) Var(2X)=4Var(X)=3, so Var(X)=3/4. So, E(3X)=Var(X/2)=Var(X)/4=3/16, so EX=1/16 (i) EX 2 =Var(X)+(EX) 2 =193/256 E[(2 + X) 2 ]=E(X 2 +4X+4)=EX 2 +4EX+4=1281/256 (ii)Var(4+3X)=9Var(X)=27/4 (b) We know Var(W)=7 E((X − W)(X + W))=E(X 2 -W 2 )=EX 2 -EW 2 =100. E(3W)=30, so EW=10. E(X+W)=EX+EW=12, so EX=2. EW 2 =VarW+(EW) 2 =7+100=107, so EX 2 =207 (i) Var(X)=EX 2 -(EX) 2 =203 (ii) cov(X 3 , W 2 )=0 (due to independence, E(X 3 W 2 )=E(X 3 )E(W 2 ).) (2). (13 points) You don’t have to calculate all the integrations. Homework is due
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This note was uploaded on 05/09/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

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HW8_solutions - HW#8 Solutions AMS 311 Probability Theory...

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