Mean and mode of a Beta random variable
A continuous random variable is said to have a $Beta(a,b)$ distribution if its density is given by
$f(x) = (1 / \text{B}(a,b))x^{a-1} (1-x)^{b-1}$ if $0 < x < 1$
Find mean , var, mode if $a = 3, b = 5.$
This is throwing me off with the beta distribution. I'm not sure if it changes the way i solve for the mean, var , mode.
My approach is that since mean of a continuous random variable which is basically the Expected value of X (EX), so we can just do $\int x * f(x) dx$. We can find $ f(x) = (105) x^{2} (1-x)^{4} $ and then $\int x *f(x) dx = 1/105 \int_{0}^{1} x^3 * (1-x)^4 dx = 105 * \beta(4,5) = 105 * ((6*24 )/ 40320)$
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$\begingroup$\begin{equation} \begin{array}{rcl} f^\prime(x)=\dfrac{df(x)}{dx}& = & \dfrac{1}{B(a,b)}\dfrac{d}{dx}x^{a-1}(1-x)^{b-1}\\ & = & \dfrac{1}{B(a,b)}\left[(a-1)x^{a-2}(1-x)^{b-1}-x^{a-1}(b-1)(1-x)^{b-2}\right]\\ & = & \dfrac{1}{B(a,b)}x^{a-2}(1-x)^{b-2}\left[(a-1)(1-x)-(b-1)x\right]\\ \end{array} \end{equation}\begin{equation} \begin{array}{rcl} f^\prime(x)=0 & \Rightarrow & (a-1)(1-x)-(b-1)x=0 \Rightarrow a-1-x(a-1+b-1)=0\\ & \Rightarrow & x=\dfrac{a-1}{a+b-2} \quad \text{exists when } \;\; a>1\;\; \& \;\;b>1 \end{array} \end{equation}
Mode exists when $a>1$ and $b>1$ and is equals to $\dfrac{a-1}{a+b-2}$
$\endgroup$ $\begingroup$Your method is OK, but you have the wrong constant term; your density function does not integrate to 1 over $(0,1).$ Look at Wikipedia for 'beta distribution'. You should get $E(X) = \alpha/(\alpha + \beta) = 3/8.$
The mode is the value of $x$ (here $x = 1/3$) at at which $f(x)$ achieves its maximum in $(0,1).$ You can find it using differential calculus.
The figure below shows the density function of this distribution. The mean is at the solid red line and the mode is at the dotted green line.
A repeat of the above, in a different manner, which I found clearer to understand:
\begin{equation} \begin{array}{rcl} f^\prime(x)=\dfrac{df(x)}{dx}& = & \dfrac{d}{dx}x^{a-1}(1-x)^{b-1}\\ & = & (a-1)x^{a-2}(1-x)^{b-1}-x^{a-1}(b-1)(1-x)^{b-2} = 0 \\ & = & x^{a-2}(1-x)^{b-2}\left[(a-1)(1-x)-(b-1)x\right] = 0 \\ & = & (a-1)(1-x)-(b-1)x = 0 \\ & = & (a-1)-x(a+b-2) = 0 \\ \end{array} \end{equation}
\begin{equation} \begin{array}{rcl} x = \dfrac{(a-1)}{(a+b-2)} = mode\\ \end{array} \end{equation}
Save my university lecturer, information taken from both the above answer, and Kaliabor College information on the Beta Distribution (p10):
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