Math Problem Statement

If f(x) = k e^{-|x|} is a probability density function in the interval, x is a real, then evaluate i) Mean ii) Variance iii) P(0 < X < 4) by finding k.

Solution

To solve the given problem, we will follow these steps systematically:


Step 1: Normalize f(x)=kexf(x) = k e^{-|x|}

To ensure f(x)f(x) is a probability density function, it must satisfy:

f(x)dx=1\int_{-\infty}^\infty f(x) \, dx = 1

We compute:

kexdx=1\int_{-\infty}^\infty k e^{-|x|} dx = 1

Since exe^{-|x|} is symmetric around x=0x = 0, the integral becomes:

2k0exdx=12k \int_{0}^\infty e^{-x} \, dx = 1

The integral of exe^{-x} from 0 to \infty is 1, so:

2k(1)=1    k=122k (1) = 1 \implies k = \frac{1}{2}

Thus, the probability density function is:

f(x)=12exf(x) = \frac{1}{2} e^{-|x|}


Step 2: Find the Mean (μ\mu)

The mean of a probability density function is:

μ=xf(x)dx\mu = \int_{-\infty}^\infty x f(x) \, dx

Due to symmetry, the mean of f(x)f(x) is 0. The positive and negative contributions to the integral cancel out:

μ=0\mu = 0


Step 3: Find the Variance (σ2\sigma^2)

The variance is given by:

σ2=x2f(x)dx\sigma^2 = \int_{-\infty}^\infty x^2 f(x) \, dx

Substituting f(x)=12exf(x) = \frac{1}{2} e^{-|x|}, we have:

σ2=12x2exdx\sigma^2 = \frac{1}{2} \int_{-\infty}^\infty x^2 e^{-|x|} \, dx

Using symmetry, we evaluate only the integral from 00 to \infty and double it:

σ2=1220x2exdx=0x2exdx\sigma^2 = \frac{1}{2} \cdot 2 \int_{0}^\infty x^2 e^{-x} \, dx = \int_{0}^\infty x^2 e^{-x} \, dx

To compute 0x2exdx\int_{0}^\infty x^2 e^{-x} \, dx, we use the gamma function (Γ(n)=(n1)!\Gamma(n) = (n-1)! for integers):

0x2exdx=Γ(3)=2!\int_{0}^\infty x^2 e^{-x} \, dx = \Gamma(3) = 2!

Γ(3)=2=2\Gamma(3) = 2 = 2

Thus:

σ2=2\sigma^2 = 2


Step 4: Find P(0<X<4)P(0 < X < 4)

To compute P(0<X<4)P(0 < X < 4), we integrate f(x)f(x) from 0 to 4:

P(0<X<4)=0412exdxP(0 < X < 4) = \int_{0}^4 \frac{1}{2} e^{-x} \, dx

The integral of exe^{-x} is ex-e^{-x}, so:

P(0<X<4)=12[ex]04P(0 < X < 4) = \frac{1}{2} \left[ -e^{-x} \right]_0^4

P(0<X<4)=12(e4(e0))P(0 < X < 4) = \frac{1}{2} \left( -e^{-4} - (-e^{0}) \right)

P(0<X<4)=12(1e4)P(0 < X < 4) = \frac{1}{2} \left( 1 - e^{-4} \right)


Final Results:

  1. Normalization constant (kk): 12\frac{1}{2}
  2. Mean (μ\mu): 00
  3. Variance (σ2\sigma^2): 22
  4. Probability P(0<X<4)P(0 < X < 4): 12(1e4)\frac{1}{2} \left( 1 - e^{-4} \right)

Would you like detailed steps on how to calculate integrals, or do you have additional questions?

Expanding Questions:

  1. How does the symmetry of a function simplify calculations for mean and variance?
  2. What is the relationship between gamma functions and factorials?
  3. Can we generalize this result for other exponential functions?
  4. What does P(0<X<4)P(0 < X < 4) represent in terms of probability?
  5. Why does exe^{-|x|} represent a valid probability distribution?

Tip:

When dealing with symmetric distributions, always check if symmetry simplifies the calculation of expected value or variance!

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Math Problem Analysis

Mathematical Concepts

Probability Density Function
Normalization
Symmetry
Expected Value
Variance
Exponential Distribution

Formulas

Integral for normalization: ∫_{-∞}^∞ f(x) dx = 1
Mean: μ = ∫_{-∞}^∞ x f(x) dx
Variance: σ² = ∫_{-∞}^∞ x² f(x) dx - μ²
Probability: P(a < X < b) = ∫_{a}^b f(x) dx

Theorems

Gamma Function Relation to Factorials
Properties of Symmetric Functions in Probability

Suitable Grade Level

Undergraduate