求连续随机变量X的概率密度$f(x)=a x\sin(x) e^{-x}$的归一化常数a

Hey there! Let's walk through solving for the normalization constant (a) step by step. I'll keep this straightforward so you can follow along easily.

Step 1: Recall the Normalization Condition #

For any continuous random variable's probability density function (PDF) (f(x)), the total area under the curve over its entire domain must equal 1. Mathematically, that means:
[
\int_{-\infty}^{\infty} f(x) dx = 1
]
In your case, (f(x) = 0) when (x \leq 0), so we can simplify the integral to just the positive real line:
[
\int_{0}^{\infty} a x \sin(x) e^{-x} dx = 1
]
We can factor out the constant (a) since it doesn't depend on (x):
[
a \cdot \int_{0}^{\infty} x \sin(x) e^{-x} dx = 1
]
Our goal now is to compute the integral (I = \int_{0}^{\infty} x \sin(x) e^{-x} dx), then solve for (a).

Step 2: Compute the Integral (I) Using Integration by Parts #

Let's break this integral down with integration by parts (remember the formula: (\int u dv = uv - \int v du)).

First, let's define:

• (u = x) (so (du = dx))
• (dv = \sin(x) e^{-x} dx) (we need to find (v) by integrating this)

Substep 2.1: Find (v = \int \sin(x) e^{-x} dx)

To compute this integral, we'll use integration by parts twice:

1. Let (u_1 = \sin(x)), (dv_1 = e^{-x} dx) → (du_1 = \cos(x) dx), (v_1 = -e^{-x})
   [
   \int e^{-x} \sin(x) dx = -e^{-x} \sin(x) + \int e^{-x} \cos(x) dx
   ]
2. Now compute (\int e^{-x} \cos(x) dx): let (u_2 = \cos(x)), (dv_2 = e^{-x} dx) → (du_2 = -\sin(x) dx), (v_2 = -e^{-x})
   [
   \int e^{-x} \cos(x) dx = -e^{-x} \cos(x) - \int e^{-x} \sin(x) dx
   ]
   Substitute this back into the first equation:
   [
   \int e^{-x} \sin(x) dx = -e^{-x} \sin(x) - e^{-x} \cos(x) - \int e^{-x} \sin(x) dx
   ]
   Bring the integral term to the left side:
   [
   2 \int e^{-x} \sin(x) dx = -e^{-x} (\sin(x) + \cos(x))
   ]
   Divide by 2:
   [
   v = \int e^{-x} \sin(x) dx = -\frac{1}{2} e^{-x} (\sin(x) + \cos(x)) + C
   ]

Substep 2.2: Apply Integration by Parts to (I)

Now plug (u), (du), (v) into the integration by parts formula for (I):
[
I = \left. uv \right|{0}^{\infty} - \int{0}^{\infty} v du
]
First evaluate (\left. uv \right|_{0}^{\infty}):

• As (x \to \infty), (e^{-x}) decays to 0 faster than (x) grows, so (uv = -\frac{1}{2} x e^{-x} (\sin(x) + \cos(x)) \to 0)
• At (x = 0), (uv = -\frac{1}{2} \cdot 0 \cdot 1 \cdot (0 + 1) = 0)
  So this term equals (0 - 0 = 0).

Now compute the remaining integral:
[
I = 0 - \int_{0}^{\infty} \left( -\frac{1}{2} e^{-x} (\sin(x) + \cos(x)) \right) dx = \frac{1}{2} \int_{0}^{\infty} e^{-x} (\sin(x) + \cos(x)) dx
]
Split this into two separate integrals:
[
I = \frac{1}{2} \left( \int_{0}^{\infty} e^{-x} \sin(x) dx + \int_{0}^{\infty} e^{-x} \cos(x) dx \right)
]
We can use standard integral formulas for these (or compute them with the same method as before):

• (\int_{0}^{\infty} e^{-x} \sin(x) dx = \frac{1}{1^2 + 1^2} = \frac{1}{2})
• (\int_{0}^{\infty} e^{-x} \cos(x) dx = \frac{1}{1^2 + 1^2} = \frac{1}{2})

Substitute these back in:
[
I = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{2} \right) = \frac{1}{2} \cdot 1 = \frac{1}{2}
]

Step 3: Solve for (a) #

Now plug (I = \frac{1}{2}) back into our normalization equation:
[
a \cdot \frac{1}{2} = 1
]
Multiply both sides by 2 to solve for (a):
[
a = 2
]

Final Check #

To confirm, if we plug (a=2) back into the original PDF, the integral over (0) to (\infty) will equal 1, which satisfies the normalization requirement for a valid PDF.

内容的提问来源于stack exchange，提问作者Fernando Martinez