Hint to Assignments 2015_02.pdf

# Hint to Assignments 2015_02.pdf - DSC3703 Assignments 01...

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DSC3703, Assignments 01 and 02: Hints/Solutions 1. Question 01: (a) Use the fact that Z + -∞ f ( x ) dx = 1 for any probability density function f ( x ). Now looking closely how f ( x ) is defined, write Z + -∞ f ( x ) dx = Z 0 -∞ f ( x ) dx + Z 2 0 f ( x ) dx + Z + 2 f ( x ) dx. By definition, f ( x ) = 0 for x ( -∞ , 0) and also f ( x ) = 0 for x (2 , + ) . Then, Z 0 -∞ f ( x ) dx = Z 0 -∞ 0 dx = 0 and Z + 2 f ( x ) dx = Z + 2 0 dx = 0 . But for x [0 , 2], we have that f ( x ) = C (3 x + x 3 ). Therefore, Z 2 0 f ( x ) dx = Z 2 0 C (3 x + x 3 ) dx = C ((3 x 2 ) / 2 + ( X 4 / 4)) 2 0 = 10 C. Now we obtain that Z + -∞ f ( x ) dx = 0 + 10 C + 0 = 10 C. Since R + -∞ f ( x ) dx = 1, then 10 C = 1, that is, C = 1 / 10. (b) To find the cumulative distribution function, use the definition, F ( x ) = Z x -∞ f ( t ) dt. Do not write Z x -∞ f ( x ) dx. Again, we have 3 cases, If x < 0 then f ( t ) = 0 for any t ( -∞ , x ] and therefore, F ( x ) = Z x -∞ f ( t ) dt = Z x -∞ 0 dt = 0 1

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If 0 x 2, then, for 0 t 2 , f ( t ) = C (3 t + t 3 ) = (1 / 10)(3 t + t 3 ) . Hence, F ( x ) = Z x -∞ f ( t ) dt = Z 0 -∞ f ( t ) dt + Z x 0 f ( t ) dt = Z 0 -∞ 0 dt + Z x 0 (1 / 10)(3 t + t 3 ) dt = (3 / 2) x 2 + (1 / 4) x 4 10 . Finally, for x > 2, we have that, F ( x ) = Z x -∞ f ( t ) dt = Z 0 -∞ f ( t ) dt + Z 2 0 f ( t ) + Z x 2 f ( t ) dt = 0 + Z 2 0 f ( t ) dt + 0 = 1 . Finally, we obtain, F ( x ) = 0 if x < 0 (3 / 2) x 2 +(1 / 4) x 4 10 if 0 x 2 1 x > 2 . We can now deduce that P { 1 / 2 X 3 / 2 } = F (3 / 2) - F (1 / 2) = 17 / 40 . 2. Question 2: a) Use again the condition Z + -∞ f ( x ) dx = 1 . Then, Z 0 -∞ f ( x ) dx + Z + 0 f ( x ) dx = 1 That is, 0 + Z + 0 Ce - 3 x dx = 1 . 2
Then Z + 0 Ce - 3 x dx = C - e - 3 x 3 + 0 = C (0 + 1 / 3) = C/ 3 We obtain that C = 3. Recall that Z e ax = e ax a . (b) The probability P { X > 1 / 3 } = Z + 1 / 3 f ( x ) dx = Z 1 / 3 3 e - 3 x dx = e - 1 0 . 3678 . 3. Question 3: The problem is to find the expectation of e αλX where X is exponentially distributed with parameter λ , i.e. X = expo ( β ) where β = 1 /λ. The density of X is given by: f ( x ) = λe - λx if x 0 0 otherwise Now the expectation of the random variable e αλX is given by: E h e αλX i = Z + -∞ f ( x ) e αλx dx. Remark: If f is the probability density function of a random variable X and g is any function, then the expectation of g ( X ) is given by E [ g ( X )] = Z + -∞ f ( x ) g ( x ) dx. Now E h e αλX i = Z + 0 λe - λx e αλx dx = Z + 0 λe ( αλ - λ ) x = λ 1 αλ - λ h e ( αλ - λ ) x i + 0 = 1 1 - α . Here note that since 0 < α < 1, then αλ - λ < 0 and hence lim x + e ( αλ - λ ) x = lim x + e - x = 0 . 3

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The final answer is P = 1 αλ ln 1 1 - α . Now replace α by 1 / 2. 4. Question 4: (a) This is very easy. Mean = E [ Y ] = X ip ( i ) = 1(1 / 2) + 2(1 / 3) + 24(1 / 6) = 31 / 6 Variance = E [ Y 2 ] - ( E [ Y ]) 2 = X i 2 p ( i ) - (31 / 6) 2 = 1 2 (1 / 2) + 2 2 (1 / 3) + 24 2 (1 / 6) - (31 / 6) 2 . (b) As indicated on page 54 of Study guide 1, if the total number of flips is 5, and the number of success (that is “head”) is 3, then the number of failure (“tails”) is 2. We know that the number of failures before the s th success follows the negative binomial distribution negbin ( s, p ). Here p = 2 / 3 and s = 3. So, X = Y + 3 where Y is the number of failures before the 3rd success ( Y = negbin (3 , 2 / 3)) Then P [ X = 5] = P [ Y = 2] = 3 + 2 + 1 2 !
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