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प्रश्न
The probability density function of the random variable X is given by
`f(x) = {{:(16x"e"^(-4x), x > 0),(0, x ≤ 0):}`
find the mean and variance of X
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उत्तर
Given p.d.f is `f(x) = {{:(16x"e"^(-4x), x > 0),(0, x ≤ 0):}`
Mean: E(X)
= `int_-oo^oo x f(x) "d"x`
= `16int_0^oo x^2 "e"^(-4x) "d"x`
Using integration by parts method twice
Let u = x2
⇒ du = 2x dx
And `int "dv" = int "e"^(-4x)`
v = `"e"^(-4x)/(-4)`
`int "u" "d" = "uv" - int "v" "du"`
`int^2"e"^(-4x) "d"x = (x^2"e"^(-4x))/(-4) + 1/4 int 2x"e"^(-4x) "d"x` .......(1)
= `- (x^2"e"^(-4x))/4 + 1/2 int x"e"^(-4x) "d"x`
∵ Integration by parts method
u = x
⇒ du = dx
And `int "dv" - int"e"^(-4x) ""x`
v = `"e"^(-4x)/(-4)`
`int "u" "dv" = "uv" - int "v" "du"`
`int x"e"^(-4x) "d"x = (-x"e"^(-4x))/4 + 1/4 int "e"^(-4x) "d"x`
= `(-x"e"^(-4x))/4 - 1/16 "e"^(-4x)`
Substituting in (1)
`intx^2"e"^(-4x) "d"x = (x^2"e"^(-4x))/(-4) + 1/2 [(-x"e"^(-4x))/4 - 1/16 e"^(-4x)]`
E(X) = `16[(x^2"e"^(-4x))/(-4) - (x"e"^(-4x))/8 - "e"^(-4x)/32]_0^oo`
= `16[0 - ((-1)/32)]`
= `16[1/32]`
= `1/2`
E(X2] = `int_-oo^oo x^2 f(x) "d"x`
= `16int_0^oo x^3"e"^(-4x) "d"x`
Using integration by parts method
Let u = x3
⇒ du = 3x2 du
And `int "dv" = int "e"^(-4x) "d"x`
⇒ v = `"e"^(-4x)/(-4)`
`int "u" "dv" = "uv" - int "v" "du"`
`intx^3"e"^(-4x) "d"x = - (x^3"e"^(-4x))/4 + 3/4 int"e"^(-4x) x^2 "d"x`
= `- (x^3"e"^(-4x))/4 + 3/4[(x^2"e"^(-4x))/(-4) - (x"e"^(-4x))/8 - "e"^(-4x)/32]`
∵ Using E(X) integration]
= `- (x^3"e"^(-4x))/4 - 3/16 x^2"e"^(-4x) - 3/32 x"e"^(-4x) - 3/128 "e"^(-4x)`
E(X2) = `16[- (x^3""^(-4x))/4 - 3/16 x^2"e"^(-4x) - 3/32 x"e"^(-4x) - 3/128 "e"^(-4x)]_0^oo`
= `16[0 - ((-3)/128)]`
= `16[3/128]`
= `3/8`
Variance Var(X) = E(X2) – [E(X)]2
= `3/8 - 1/4`
= `(3 - 2)/8`
= `1/8`
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