The mode of beta distribution

[Double T]

hi all,

I have a question,

A random variable has a beta distribution with parameters a>1 and b>1.
Determine the mode.

the answer is (a-1)/(a+b-2)

Why? how to find it??

Thanks for all your help

T

[Greg1983]

Originally Posted by Double T

hi all,

I have a question,

A random variable has a beta distribution with parameters a>1 and b>1.
Determine the mode.

the answer is (a-1)/(a+b-2)

Why? how to find it??

Thanks for all your help

T

This looked interesting so I had to try it. The mode is where the pdf reaches its maximum, so differentiate the pdf and set it equal to zero. The pdf is given by:
(Gamma(a+b)/[Gamma(a)*Gamma(b)]) * x^(a-1) * (1-x)^(b-1)
Differentiating yields via product rule (and setting equal to zero):
(Gamma(a+b)/[Gamma(a)*Gamma(b)]) * [(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2)] = 0
Simplify:
(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2) = 0
(a-1)*(1-x) - x*(b-1) = 0
a - ax - 1 + x - xb + x = 0
x(2 - a - b) = 1 - a
x = (1-a)/(2-a-b) = (a-1)/(a+b-2)

Hopefully that all looks good to you!

[Double T]

Originally Posted by Greg1983

This looked interesting so I had to try it. The mode is where the pdf reaches its maximum, so differentiate the pdf and set it equal to zero. The pdf is given by:
(Gamma(a+b)/[Gamma(a)*Gamma(b)]) * x^(a-1) * (1-x)^(b-1)
Differentiating yields via product rule (and setting equal to zero):
(Gamma(a+b)/[Gamma(a)*Gamma(b)]) * [(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2)] = 0
Simplify:
(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2) = 0
(a-1)*(1-x) - x*(b-1) = 0
a - ax - 1 + x - xb + x = 0
x(2 - a - b) = 1 - a
x = (1-a)/(2-a-b) = (a-1)/(a+b-2)

Hopefully that all looks good to you!

Thank you Greg,

But i have one more question, how about (Gamma(a+b)/[Gamma(a)*Gamma(b)])??
We don't need to differentiate this part??

Thanks,
T

[Greg1983]

Originally Posted by Double T

Thank you Greg,

But i have one more question, how about (Gamma(a+b)/[Gamma(a)*Gamma(b)])??
We don't need to differentiate this part??

Thanks,
T

No because you're differentiating with respect to x. It stays out the front as a constant then you can just get rid of it when you set the whole thing equal to zero.

[Double T]

Originally Posted by Greg1983

No because you're differentiating with respect to x. It stays out the front as a constant then you can just get rid of it when you set the whole thing equal to zero.

''', how come i didn't think about it??

Great, thank you very much, Greg..........