Chapter 5: Q. 5.18 (page 216)

Verify that the gamma density function integrates to $1$ .

Short Answer

Expert verified

To establish the assertion, integrate $F$ over $ℝ$ and analyze the definition of the Gamma function.

Step by step solution

Find The density function.

Suppose that $X ~$ Gamma $(α, λ)$ The density function is

$f (x) = \frac{λ^{α}}{Γ (α)} x^{α - 1} e^{- λ x}$

For $x \geq 0$ otherwise it is equal to zero. We want to prove that

$\int_{ℝ} f (x) d x = 1$

We have that,

localid="1649661055597" $\int_{ℝ} f (x) d x = \int_{0}^{\infty} \frac{λ^{α}}{Γ (α)} x^{α - 1} e^{- λ x} d x$

= \frac{λ^{α}}{Γ (α)} \int_{0}^{\infty} x^{α - 1} e^{- λ x} d x

(1)

Evaluate the integral process.

In order to evaluate the integral, make the substitution $s = λ x$ that yields $x = \frac{s}{λ}$ and $d s = λ d x$ we have that

localid="1649661029116" $\int_{0}^{\infty} x^{α - 1} e^{- λ x} d x = \int_{0}^{\infty} {(\frac{s}{λ})}^{α - 1} e^{- s} \frac{d s}{λ}$

$= \frac{1}{λ^{α}} \int_{0}^{\infty} s^{α - 1} e^{- s} d s = \frac{Γ (α)}{λ^{α}}$

If that plug $(1)$ , we end up with

localid="1649661004300" $\frac{λ^{α}}{Γ (α)} \int_{0}^{\infty} x^{α - 1} e^{- λ x} d x$

$= \frac{λ^{α}}{Γ (α)} \times \frac{Γ (α)}{λ^{α}} = 1$

Hence, we have proved the claim.

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Verify that the gamma density function integrates to $1$ .

Short Answer

Step by step solution

Find The density function.

Evaluate the integral process.

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Verify that the gamma density function integrates to1.

Short Answer

Step by step solution

Find The density function.

Evaluate the integral process.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Pure Maths

Discrete Mathematics

Mechanics Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Verify that the gamma density function integrates to $1$ .