Definition of Beta Type I Distribution
A continuous random variable X is said to have a beta type I distribution with parameters α and β if its p.d.f. is given by
f(x)={1B(α,β)xα−1(1−x)β−1,0≤x≤1;0,Otherwise.
where,
-
B(α,β)=ΓαΓβΓ(α+β)=∫10xα−1(1−x)β−1dx
is a beta function and -
Γ(α)
is a gamma function.
Mean of Beta Type I Distribution
The mean of beta type I distribution is E(X)=αα+β.
Variance of Beta Type I Distribution
The variance of beta type I distribution is V(X)=αβ(α+β)2(α+β+1).
Example 1
Suppose the proportion X of surface area in a randomly selected quadrant that is covered by a certain plant has a beta distribution with α=5 and β=2.
Calculate
a. E(X) and V(X),
b. P(X≤0.2)
c. P(0.2≤X≤0.4)
Solution
Let X denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that X∼β1(α,β).
The probability density function of X is
f(x)=1B(5,2)x5−1(1−x)2−1;0≤x≤1=Γ(5+2)Γ(5)Γ(2)x4(1−x)1;0≤x≤1=6!4!1!x4(1−x)1;0≤x≤1=30x4(1−x);0≤x≤1
a. Mean and variance of X
E(X)=αα+β=55+2=0.7143.
V(X)=αβ(α+β)2(α+β+1)=5×2(5+2)2(5+2+1)=0.0255
b. P(X≤0.2)
P(X≤0.2)=∫0.20f(x)dx=30∫0.20x4(1−x)dx=30∫0.20(x4−x5)dx=30[x55−x66]0.20=30[0.255−0.266]=0.0016
c. P(0.2≤X≤0.4)
P(0.2≤X≤0.4)=∫0.40.2f(x)dx=30∫0.40.2x4(1−x)dx=30∫0.40.2(x4−x5)dx=30[x55−x66]0.40.2=30[0.455−0.466−0.255+0.266]=0.0394
Example 2
The proportion of a brand of a television set that requires repair during the first year is a random variable having a beta distribution with α=3,β=2.
a. Find the average and standard deviation of proportion of a brand of a television set that requires repair during the first year.
b. What is the probability that at least 80% of the new models sold this year will require repairs during the first year of operation?
c. What is the probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation?
Solution
Let X denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that X∼β1(α,β).
The probability density function of X is
f(x)=1B(3,2)x3−1(1−x)2−1;0≤x≤1=Γ(3+2)Γ(3)Γ(2)x2(1−x)1;0≤x≤1=4!2!1!x2(1−x)1;0≤x≤1=12x2(1−x);0≤x≤1
a. Mean of X is
E(X)=αα+β=33+2=0.6.
Variance of X is
V(X)=αβ(α+β)2(α+β+1)=3×2(3+2)2(3+2+1)=0.04.
Thus the standard deviation of X is
σ=√V(X)=√0.04=0.2.
b. The probability that at least 80% of the new models sold this year will require repairs during the first year of operation i.e., P(X≥0.8)
P(X≥0.8)=∫10.8f(x)dx=12∫10.8x2(1−x)dx=12∫10.8(x2−x3)dx=12[x33−x44]10.8=12[133−144−0.833+0.844]=0.1808
c. The probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation i.e., P(0.6≤X≤0.8)
P(0.6≤X≤0.8)=∫0.80.6f(x)dx=12∫0.80.6x2(1−x)dx=12∫0.80.6(x2−x3)dx=12[x33−x44]0.80.6=12[0.833−0.844−0.633+0.644]=0.344
I hope this article helps you understand how to solve the numerical problems on Beta type I distribution.