Chapter 5: Q. 5.26 (page 216)

If $x$ is a beta random variable with parameters $a$ and $b$ , show that
$E [X] = \frac{a}{a + b}$
$Var (X) = \frac{ab}{(a + b)^{2} (a + b + 1)}$

Short Answer

Expert verified

In general, find the $k$ th moment and use it to calculate the mean and variance.

Step by step solution

Beta distribution

Density function,

$f (x) = \frac{Γ (α + β)}{Γ (α) Γ (β)} \times x^{α - 1} (1 - x)^{β - 1}$

$x \in [0, 1]$ is to be zero.

$E (X^{k}) = \int_{ℝ} x^{k} f (x) dx = \frac{Γ (α + β)}{Γ (α) Γ (β)} \int_{0}^{1} x^{k} \times x^{α - 1} (1 - x)^{β - 1} dx$

$= \frac{Γ (α + β)}{Γ (α) Γ (β)} \int_{0}^{1} x^{k + α - 1} (1 - x)^{β - 1} dx$

So,

localid="1649485821665" $E (X^{k}) = \frac{Γ (α + β)}{Γ (α) Γ (β)} \times B (k + α, β)$

$= \frac{Γ (α + β)}{Γ (α) Γ (β)} \times \frac{Γ (k + α) Γ (β)}{Γ (k + α + β)}$

Explanation

For $k = 1$

$E (X) = \frac{Γ (α + β)}{Γ (α)} \times \frac{Γ (α + 1)}{Γ (α + β + 1)}$

$= \frac{Γ (α + β)}{Γ (α)} \times \frac{αΓ (α)}{(α + β) Γ (α + β)} = \frac{α}{α + β}$

For $k = 2$

$E (X^{2}) = \frac{Γ (α + β)}{Γ (α)} \times \frac{Γ (α + 2)}{Γ (α + β + 2)}$

$= \frac{Γ (α + β)}{Γ (α)} \times \frac{(α + 1) αΓ (α)}{(α + β + 1) (α + β) Γ (α + β)}$

So,

$Var (X) = E (X^{2}) - E (X)^{2}$

$= \frac{(α + 1) α}{(α + β + 1) (α + β)} - \frac{α^{2}}{(α + β)^{2}}$

$= \frac{(α + 1) α (α + β) - α^{2} (α + β + 1)}{(α + β + 1) (α + β)^{2}}$

$= \frac{αβ}{(α + β + 1) (α + β)^{2}}$

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If $x$ is a beta random variable with parameters $a$ and $b$ , show that
$E [X] = \frac{a}{a + b}$
$Var (X) = \frac{ab}{(a + b)^{2} (a + b + 1)}$

Short Answer

Step by step solution

Beta distribution

Explanation

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If xis a beta random variable with parameters aand b, show thatE[X]=aa+bVar(X)=ab(a+b)2(a+b+1)

Short Answer

Step by step solution

Beta distribution

Explanation

One App. One Place for Learning.

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If $x$ is a beta random variable with parameters $a$ and $b$ , show that
$E [X] = \frac{a}{a + b}$
$Var (X) = \frac{ab}{(a + b)^{2} (a + b + 1)}$