Normalizing a PDF

Probability Level 2

A certain continuous random variable has a probability density function (PDF) given by:

f ( x ) = C x ( 1 − x ) 2 , f(x) = C x (1-x)^2,

where x x can be any number in the real interval [ 0 , 1 ] [0,1] . Compute C C using the normalization condition on PDFs.

2 12 24 6

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Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
May 10, 2016

The normalization condition is the integral:

∫ 0 1 C x ( 1 − x ) 2 d x = 1. \int_0^1 Cx(1-x)^2 dx = 1.

Computing using integration by parts,

∫ 0 1 C x ( 1 − x ) 2 d x = − 1 3 C x ( 1 − x ) 3 ∣ 0 1 + 1 3 ∫ 0 1 C ( 1 − x ) 3 d x = 1 3 ∫ 0 1 C ( 1 − x ) 3 d x = − 1 3 1 4 C ( 1 − x ) 4 ∣ 0 1 = C 12 . \begin{aligned} \int_0^1 Cx(1-x)^2 dx &= \bigl.-\frac13 Cx(1-x)^3\bigr|_0^1 + \frac13 \int_0^1 C(1-x)^3 dx \\ &= \frac13 \int_0^1 C(1-x)^3 dx = -\bigl.\frac13 \frac14 C (1-x)^4 \bigr|_0^1 = \frac{C}{12}. \end{aligned}

Thus C = 12 C=12 by the normalization condition.

18 Helpful 6 Interesting 4 Brilliant 22 Confused

Why did you do an integration by parts? It doesn't make sense to me. It is rather simple to solve this integral using this:

∫ 0 1 C x ( 1 − x ) 2 d x = C ⋅ ∫ 0 1 x − 2 x 2 + 2 x 3 . . . \int_0^1 Cx(1-x)^2 dx = C \cdot \int_0^1 x - 2x^2 + 2x^3 ...

Just apply the power rule and it would return the result 1 12 \frac{1}{12} while yours result seems strange to me, would you care to explain more?

Rafael Nunes - 2 years, 10 months ago

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"Just apply the power rule" is also pretty much what I did. You had to integrate three terms and do some sums and differences, I only had to integrate one term after doing the integration by parts ;)

The integration by parts was convenient for me because I could see beforehand that the boundary (first) term would evaluate to zero at both endpoints. Honestly, either method seems equally elementary to me.

Matt DeCross - 2 years, 9 months ago

how did you get 1/3?

Chardylaine Jocson - 2 years, 3 months ago

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That came from integrating the (1-x)^2 part of the integrand and then he just moved the 1/3 outside the integral since it is a constant factor.

Tristan Goodman - 2 years, 2 months ago

It’s absolutely amazing to me that anyone can understand this, lol

Vinny Loglio - 1 year, 1 month ago

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its not 'that' difficult. Integration by parts is a simple technique. There are vastly more complicated things on this site then this.

Hafiz muhammad Ibrahim jaffar - 11 months, 1 week ago

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