Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.

Let $A$ be a Beta distributed random variable, $\mathcal B(a,b)$, with probability density function: $$ f(x)= \frac{Γ(a+b)}{Γ(a)Γ(b)}*x^{(a-1)}*(1-x)^{(b-1)}, $$ where $0<x<1$. Suppose that $B$ is a random variable, $\mathcal B(a+1,b)$. Write down its probability density function $g(x)$.

share|improve this question
2  
Hint: Just replace $a$ by $a+1$ in $f(x)$. – user10525 Aug 14 '12 at 15:43
2  
Welcome to the site, @GeorgiaMarkadji. I edited your question w/ $\TeX$ to make the equations clearer; make sure it says what you want it to. I also took the liberty of adding the homework tag, as this question sounds like a class assignment, or that it came from a book; feel free to comment or change that back if it's not true. Note that, in keeping w/ our FAQ we don't usually answer homework questions directly, but provide hints so that you can figure it out for yourself. – gung Aug 14 '12 at 15:45
what is the relationship between the two functions f(x) and g(x)? – Georgia Markadji Aug 14 '12 at 16:03
@GeorgiaMarkadji You can see this if you add a bit of notation, say $f(x;a,b)=\frac{Γ(a+b)}{Γ(a)Γ(b)}*x^{(a-1)}*(1-x)^{(b-1)}$ is the probability density function of a Beta distributed random variable with parameters $a$ and $b$. Then if you want the probability density function of a Beta random variable with parameters $a+1$ and $b$ you just have to replace $a$ by $a+1$, this is, $f(x;a+1,b)$. Therefore the relationship between $g$ and $f$ is $g(x)=f(x;a+1,b)$. Is this clear? – user10525 Aug 14 '12 at 16:23
@Procrastinator Yes, it is very clear! thank you very much for your help. – Georgia Markadji Aug 14 '12 at 16:26
show 4 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.