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Question Determine the constant kk that makes the given density function f(x)f(x) a valid probability density function for the measurement error XX.

Studdy Solution

STEP 1

Assumptions
1. f(x) f(x) is a probability density function (pdf).
2. A valid pdf must satisfy the property that the integral of f(x) f(x) over the entire space must equal 1.
3. The given function f(x) f(x) is defined piecewise with a quadratic term within the interval [−1,1][-1, 1] and is 0 elsewhere.

STEP 2

To find the value of k k that makes f(x) f(x) a valid density function, we must set up the integral of f(x) f(x) over the entire space and solve for k k .
∫−∞∞f(x) dx=1 \int_{-\infty}^{\infty} f(x) \, dx = 1

STEP 3

Since f(x)=0 f(x) = 0 for x x outside the interval [−1,1][-1, 1], we only need to integrate f(x) f(x) from −1-1 to 11.
∫−11k(4−x2) dx=1 \int_{-1}^{1} k(4 - x^2) \, dx = 1

STEP 4

Now, we compute the integral.
∫−11k(4−x2) dx=k∫−11(4−x2) dx \int_{-1}^{1} k(4 - x^2) \, dx = k \int_{-1}^{1} (4 - x^2) \, dx

STEP 5

We can split the integral into two parts.
k∫−11(4−x2) dx=k(∫−114 dx−∫−11x2 dx) k \int_{-1}^{1} (4 - x^2) \, dx = k \left( \int_{-1}^{1} 4 \, dx - \int_{-1}^{1} x^2 \, dx \right)

STEP 6

Calculate the first integral.
∫−114 dx=4x∣−11=4(1)−4(−1)=8 \int_{-1}^{1} 4 \, dx = 4x \Big|_{-1}^{1} = 4(1) - 4(-1) = 8

STEP 7

Calculate the second integral, which is symmetric around 0 and thus can be computed over the half interval and then doubled.
∫−11x2 dx=2∫01x2 dx \int_{-1}^{1} x^2 \, dx = 2 \int_{0}^{1} x^2 \, dx

STEP 8

Compute the integral of x2 x^2 from 0 to 1.
∫01x2 dx=x33∣01=133−033=13 \int_{0}^{1} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}

STEP 9

Double the result of the integral from 0 to 1 to account for the interval from -1 to 1.
2∫01x2 dx=2⋅13=23 2 \int_{0}^{1} x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}

STEP 10

Substitute the results of the integrals back into the equation.
k(8−23)=k(243−23)=k⋅223 k(8 - \frac{2}{3}) = k \left( \frac{24}{3} - \frac{2}{3} \right) = k \cdot \frac{22}{3}

STEP 11

Set the result equal to 1 and solve for k k .
k⋅223=1 k \cdot \frac{22}{3} = 1

STEP 12

Divide both sides by 223 \frac{22}{3} to isolate k k .
k=1223 k = \frac{1}{\frac{22}{3}}

STEP 13

Invert the fraction on the right-hand side to solve for k k .
k=322 k = \frac{3}{22}
k k must be 322 \frac{3}{22} to make f(x) f(x) a valid density function.

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