Solved on Feb 22, 2024

Determine the constant $k$ that makes the given density function $f(x)$ a valid probability density function for the measurement error $X$ .

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

Calculate the first integral.
$\int_{-1}^{1} 4 \, dx = 4x \Big|_{-1}^{1} = 4(1) - 4(-1) = 8$

STEP 7

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

STEP 8

Compute the integral of $x^2$ from 0 to 1.
$\int_{0}^{1} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

STEP 9

Double the result of the integral from 0 to 1 to account for the interval from -1 to 1.
$2 \int_{0}^{1} x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

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Determine the constant $k$ that makes the given density function $f(x)$ a valid probability density function for the measurement error $X$ .

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

Calculate the first integral.
$\int_{-1}^{1} 4 \, dx = 4x \Big|_{-1}^{1} = 4(1) - 4(-1) = 8$

STEP 7

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

STEP 8

Compute the integral of $x^2$ from 0 to 1.
$\int_{0}^{1} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

STEP 9

Double the result of the integral from 0 to 1 to account for the interval from -1 to 1.
$2 \int_{0}^{1} x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

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Determine the constant kkk that makes the given density function f(x)f(x)f(x) a valid probability density function for the measurement error XXX.

STEP 1

STEP 2

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

STEP 3

Since f(x)=0 f(x) = 0 f(x)=0 for x x x outside the interval [−1,1][-1, 1][−1,1], we only need to integrate f(x) f(x) f(x) from −1-1−1 to 111.∫−11k(4−x2) dx=1 \int_{-1}^{1} k(4 - x^2) \, dx = 1 ∫−11​k(4−x2)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 ∫−11​k(4−x2)dx=k∫−11​(4−x2)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) k∫−11​(4−x2)dx=k(∫−11​4dx−∫−11​x2dx)

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 ∫−11​4dx=4x∣∣​−11​=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 ∫−11​x2dx=2∫01​x2dx

STEP 8

Compute the integral of x2 x^2 x2 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} ∫01​x2dx=3x3​∣∣​01​=313​−303​=31​

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} 2∫01​x2dx=2⋅31​=32​

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} k(8−32​)=k(324​−32​)=k⋅322​

STEP 11

Set the result equal to 1 and solve for k k k.k⋅223=1 k \cdot \frac{22}{3} = 1 k⋅322​=1

STEP 12

Divide both sides by 223 \frac{22}{3} 322​ to isolate k k k.k=1223 k = \frac{1}{\frac{22}{3}} k=322​1​

STEP 13

Invert the fraction on the right-hand side to solve for k k k.k=322 k = \frac{3}{22} k=223​k k k must be 322 \frac{3}{22} 223​ to make f(x) f(x) f(x) a valid density function.

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Determine the constant $k$ that makes the given density function $f(x)$ a valid probability density function for the measurement error $X$ .

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

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

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

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

Calculate the first integral.
$\int_{-1}^{1} 4 \, dx = 4x \Big|_{-1}^{1} = 4(1) - 4(-1) = 8$

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

Compute the integral of $x^2$ from 0 to 1.
$\int_{0}^{1} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

Double the result of the integral from 0 to 1 to account for the interval from -1 to 1.
$2 \int_{0}^{1} x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$

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

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

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

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