If a r.v. X has p.d.f., f (x) = cx , for 1 < x < 3, c > 0, Find c, E(X) and Var (X). - Mathematics and Statistics

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Sum

If a r.v. X has p.d.f., 

f (x) = `c /x` , for 1 < x < 3, c > 0, Find c, E(X) and Var (X).

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Solution

Since, f(x) is p.d.f. of r.v. X

∴` int_(-∞)^∞ f (x) dx` = 1

∴` int_(-∞)^1 f (x) dx + int_(3)^1f(x) dx+ int_(3)^∞f(x) dx = 1`

∴ `0 +int_(1)^3 f (x) dx +0 = 1`

∴ `int_(1)^3 c/x dx = 1`

∴ `c int_(1)^3 1/x dx = 1`

∴ `c [log x]_1^3 = 1`

∴ c [log 3 - log 1] = 1

∴ `1/log 3` ...........[∵ log 1 =  0]

E(X) = ` int_(-∞)^∞x f (x) dx = int_(-∞)^1x f (x) dx + int_(1)^3x f (x) dx + int_(3)^∞x f (x) dx`

`= 0 + int_(1)^3x f (x) dx + 0 = int_(1)^3x . c/x dx`

= `c int_(1)^31dx , where  c = 1/log3`

= `1/log3[x]_1^3 = 1/log 3[ 3-1] = 2/log3`

= consider, ` int_(-∞)^∞ x^2f (x) dx = int_(-∞)^1 x^2f (x) dx +int_(1)^3 x^2f (x) dx + int_(3)^∞ x^2f (x) dx`

= 0 +  `int_(1)^3 x^2f (x) dx + 0 = int_(-∞)^∞ x^2. c/x dx `

= `1/log3 int_(1)^3 x  dx = 1/log 3[x^2/2]_1^3`

= `1/log3[9/2-1/2] = 4/log3`

Now, var (x) = `int_(-∞)^∞x^2 f (x) dx - [ E (x)] ^2`

= `4/log3 - (2/log3)^2`

= `4/log3 - 4/(log3)^2`

= `4 (log3) - 4/(log3)^2 = (4[log3-1])/(log3)^2`

Hence, `c =1/log3, E(x) = 2/log3 and Var (x) = (4[log3-1])/(log3)^2`

Concept: Probability Distribution of Discrete Random Variables
  Is there an error in this question or solution?
Chapter 7: Probability Distributions - Exercise 7.2 [Page 239]

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