丁香园AV

Verify that \(f\) is a probability density function on the given interval.

1. \(f(x) = \dfrac{10}{3x^2}, x \in [2,5]\)

   Solution
   Since \(f(x) > 0\) for all \(x \in[2,5]\text{,}\) \(f\) satisfies the positivity condition. We now calculate the area under the curve of \(f\text{:}\)

   \begin{equation*} \int_2^5 \frac{10}{3x^2}\,dx = \frac{10}{3} \left[-\frac{1}{x}\right]_2^5 = \frac{10}{3}\frac{3}{10} = 1. \end{equation*}

   Thus, \(f\) is a probability density function on \([2,5]\text{.}\)

2. \(f(x) = 6\left(\sqrt{x}-x\right), x \in [0,1]\)

   Solution

   Since \(\sqrt{x} \geq x\) on \([0,1]\text{,}\) we have that \(\sqrt{x} - x \geq 0\) and so \(f\) satisfies the positivity condition. We now calculate the area under the curve of \(f\text{:}\)

   \begin{equation*} \int_0^1 6(\sqrt{x}-x)\,dx = 6\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^1 = 6\left[\frac{2}{3}-\frac{1}{2}\right]=1 \end{equation*}

   Thus, \(f\) is a probability density function on \([0,1]\text{.}\)

3. \(f(x,y) = \dfrac{1}{3}, (x,y)\in[1,2]\times[3,6]\)

   Solution
   Clearly, \(f\) satisfies the positivity condition. We now integrate:

   \begin{equation*} \int_1^2 \int_3^6 \frac{1}{3} \,dy\,dx = \frac{1}{3} \int_1^2 3\,dx = \frac{6-3}{3} = 1. \end{equation*}

   Hence, \(f\) is a probability density function on \([1,2]\times[3,6]\text{.}\)

4. \(f(x,y) = \dfrac{1}{4} xy, (x,y) \in[0,1]\times[0,4]\)

   Solution

   We see that \(f(x,y) \geq 0\) for all \(x,y \in [0,1]\times[0,4]\text{.}\) Next, we calculate

   \begin{equation*} \int_0^1\int_0^4 \frac{1}{4}xy\,dy\,dx = \int_0^1 \frac{1}{8}xy^2\bigg\vert_0^4\,dx = \int_0^1 2x\,dx = 1\text{.} \end{equation*}

   Hence, \(f(x,y)\) is a probability density function on \([0,1]\times[0,4]\text{.}\)

Construct a probability density function \(f\) from the given function \(g\text{.}\)

1. \(g(x)=3-x, x\in[0,3]\)

   Answer
   \(f(x)=\frac{2}{9}(3-x), x\in[0,3]\)
   Solution
   We integrate \(g(x)\) over \([0,3]\text{:}\)

   \begin{equation*} \int_0^3 (3-x)\,dx = \left[3x-\frac{x^2}{2}\right]_0^3 = \frac{9}{2}. \end{equation*}

   So let \(f(x) = \frac{2}{9} (3-x)\text{.}\) Then we notice that \(f(x) \geq 0\) on \([0,3]\text{.}\) Hence, \(f\) is a probability density function on \([0,3]\text{.}\)

2. \(g(x)=\dfrac{1}{x^5}, x \in[1,\infty)\)

   Answer
   \(f(x)= \frac{4}{x^5}, x\in[1,\infty)\)
   Solution

   We compute

   \begin{equation*} \begin{split} \int_1^{\infty} g(x)\,dx \amp = \int_1^{\infty} \frac{1}{x^5}\,dx \\ \amp = \lim_{a\to\infty} \int_1^{a} \frac{1}{x^5}\,dx \\ \amp = \lim_{a\to\infty} \left.-\frac{1}{4x^4} \right\vert_1^a\\ \amp = \left(0+\frac{1}{4}\right) = \frac{1}{4} \end{split} \end{equation*}

   Therefore, we take \(f(x)= \dfrac{4}{x^5}\text{.}\) Clearly, \(f(x) \geq 0\) on \([1,\infty]\text{,}\) and so \(f(x) = \dfrac{4}{x^5}\) is a probability density function on \([1,\infty]\text{.}\)

3. \(g(x,y) = xy^2, (x,y)\in [0,1]\times[0,1]\)

   Answer
   \(f(x,y)=6xy^2, (x,y)\in[0,1]\times[0,1]\)
   Solution

   We integrate \(g(x,y)\) over \([0,1]\times[0,1]\text{:}\)

   \begin{equation*} \int_0^1\int_0^1 xy^2\,dx\,dy = \int_0^1 \frac{1}{2}y^2\,dy = \frac{1}{6}y^3\bigg\vert_0^1 = \frac{1}{6} \end{equation*}

   Therefore, we take \(f(x,y) = 6xy^2\text{,}\) and we notice that \(f(x,y) \geq 0\) on the given interval. Thus, \(f(x,y)= 6xy^2\) is a probability density function on \([0,1]\times[0,1]\text{.}\)

4. \(g(x,y) = x^2e^{-y}, (x,y)\in [1,2]\times[1,\infty)\)

   Answer
   \(f(x,y)=\frac{3e}{7}x^2e^{-y}, (x,y)\in [1,2]\times[1,\infty)\)
   Solution
   We first integrate \(g(x,y)\) over the given region:

   \begin{equation*} \begin{split} \int_1^{\infty} \int_1^2 x^2 e^{-y} \,dx\,dy \amp= \frac{7}{3} \int_1^{\infty} e^{-y}\,dy\\ \amp= \frac{7}{9} \left[e^{-1}-\lim_{y\to\infty} e^{-y}\right]\\ \amp= \frac{7}{9e}.\end{split} \end{equation*}

   Therefore, let \(f(x,y) = \frac{7}{9e} x^2 e^{-y} = \frac{7}{9} x^2 e^{-(y+1)}\) (notice that \(f(x,y) \geq 0\)). Then \(f\) is a probability density function on the given interval.

Calculate the following cumulative distributions.

1. \(f(x)=\dfrac{1}{2}e^{-x/2}, x\in[0,\infty)\)

   1. \(P(x=1)\)

   2. \(P(3\leq x \leq 6)\)

   3. \(P(x \leq 50)\)

   4. \(P(x \geq 6)\)

   Answer

   i.0, ii. 0.173, iii.\(\approx 1\text{,}\) iv. 0.0498

   Solution

   Given the probability density function \(f(x)=\dfrac{1}{2}e^{-x/2}\) for \(x \in [0,\infty)\text{,}\) we calculate the following probabilities:

   1. \(P(x=1) = 0\) since \(f\) is a continuous probability distribution.

   2. \(\begin{aligned}P(3\leq x \leq 6) \amp = \int_3^6 f(x)\,dx = \int_3^6 \frac{1}{2}e^{-x/2}\,dx\\ \amp = \frac{1}{2}\left(-2 e^{-x/2} \big\vert_3^6\right)\\\amp = e^{-3/2}-e^{-3} \approx 0.173 \end{aligned}\)

   3. \(\begin{aligned}P(x \leq 50) \amp = \int_{-\infty}^{50} f(x)\,dx \\ \amp= \int_0^{50} \frac{1}{2}e^{-x/2}\,dx\\ \amp = -e^{-x/2}\big\vert_0^{50} = 1-\frac{1}{e^{25}} \approx 1 \end{aligned}\)

   4. \(\begin{aligned}P(x \geq 6) \amp = \int_6^{\infty} f(x)\,dx\\ \amp= \int_6^{\infty} \frac{1}{2}e^{-x/2}\,dx\\ \amp= \lim_{a \to \infty} \int_6^a \frac{1}{2}e^{-x/2}\,dx \\ \amp = -\lim_{a\to\infty} e^{-x/2}\big\vert_6^a = e^{-3}-\lim_{a\to\infty}e^{-a/2} = e^{-3} \approx 0.05 \end{aligned}\)

2. \(f(x)=\dfrac{3}{14}\sqrt{x}, x \in[1,4]\)

   1. \(P(2 \leq x \leq 4)\)

   2. \(P(1\leq x \leq 3)\)

   3. \(P(x \leq 2)\)

   4. \(P(x \geq 2)\)

      Answer

      i. 0.739, ii. 0.261, iii.0.599, iv. 0.739

      Solution
      Given the probability density function \(f(x)=\dfrac{3}{14}\sqrt{x}\) for \(x \in [1,4]\text{,}\) we calculate the following probabilities:

      1. \begin{equation*} \begin{aligned} P(2 \leq x \leq 4) = \int_2^4 f(x)\,dx = \frac{3}{14}\int_2^4 \sqrt{x}\,dx \amp= \frac{3}{14} \frac{2}{3} x^{3/2}\big\vert_2^4 \\ \amp= \frac{3}{14}\frac{2}{3}\left(4^{3/2} - 2^{3/2}\right) \approx 0.73880. \end{aligned} \end{equation*}
      2. \begin{equation*} \begin{aligned} P(1 \leq x \leq 3) = \int_1^3 f(x)\,dx = \frac{3}{14}\int_1^3 \sqrt{x}\,dx \amp= \frac{3}{14} \frac{2}{3} x^{3/2}\big\vert_1^3 \\ \amp= \frac{3}{14}\frac{2}{3}\left(3^{3/2} - 1\right) \approx 0.59945.\end{aligned} \end{equation*}
      3. \begin{equation*} \begin{aligned} P(x \leq 2) = \int_1^2 f(x)\,dx = \frac{3}{14}\int_1^2 \sqrt{x}\,dx \amp= \frac{3}{14} \frac{2}{3} x^{3/2}\big\vert_1^2 \\ \amp= \frac{3}{14}\frac{2}{3}\left(2^{3/2} - 1\right) \approx 0.26120.\end{aligned} \end{equation*}
      4. \begin{equation*} \begin{aligned}[t] P(x \geq 2) = \int_2^4 f(x)\,dx = \frac{3}{14}\int_2^4 \sqrt{x}\,dx \amp= \frac{3}{14} \frac{2}{3} x^{3/2}\big\vert_2^4 \\ \amp= \frac{3}{14}\frac{2}{3}\left(4^{3/2} - 2^{3/2}\right) \approx 0.73880.\end{aligned} \end{equation*}

3. \(f(x,y)=\dfrac{1}{3}xy, (x,y)\in[0,2]\times[1,2]\)

   1. \(P(0\leq x\leq 1, 1\leq y \leq 2)\)

   2. \(P(1 \leq x \leq 2, y = 1)\)

   Answer

   i. 0.25, ii. 0

   Solution

   Given the joint probability density function \(f(x,y)=\dfrac{1}{3}xy\) on \([0,2]\times[1,2]\text{,}\) we calculate the following probabilities:

   1. \(\begin{aligned}P(0\leq x\leq 1, 1 \leq y \leq 2) \amp = \int_0^1 \int_1^2 \frac{1}{3}xy\,dy\,dx \\ \amp = \int_0^1 \left.\frac{1}{6} xy^2\right\vert_1^2\,dx \\ \amp = \int_0^1 \frac{1}{3} x \,dx \\ \amp = \left.\frac{1}{6}x^2\right\vert_0^1 = \frac{1}{6} \end{aligned}\)

   2. \(\begin{aligned}P(1 \leq x \leq 2, y = 1) \amp = \int_1^2 \int_1^1 \frac{1}{3}xy\,dy \,dx \\ \amp = \int_1^2 0 \,dx = 0 \end{aligned}\)

4. \(f(x,y)=\dfrac{1}{16}(2-x)y, (x,y)\in[0,2]\times[0,4]\)

   1. \(P(0 \leq x \leq 1, 0 \leq y \leq 1)\)

   2. \(P(x \geq 1, y \geq 3)\)

   Answer

   i. 0.0469, ii. 0.109

   Solution
   Given the joint probability density function \(f(x,y)=\dfrac{1}{16}(2-x)y\) on \([0,2]\times[0,4]\text{,}\) we calculate the following probabilities:

   1. \begin{equation*} \begin{aligned} P(0\leq x\leq 1, 0 \leq y \leq 1) \amp= \int_0^1 \int_0^1 \dfrac{1}{16}(2-x)y \,dy\,dx \\ \amp= \frac{3}{64} \approx 0.046875. \end{aligned} \end{equation*}
   2. \begin{equation*} \begin{aligned} P(x \geq 1, y \geq 3) \amp= \int_1^2 \int_3^4 \dfrac{1}{16}(2-x)y \,dy\,dx\\ \amp= \frac{7}{64} \approx 0.109375. \end{aligned} \end{equation*}

Let

Show that \(f\) is a probability density function, and that the distribution has no mean.

Let

Show that \(\ds \int_{-\infty }^\infty f(x)\,dx = 1\text{.}\) Is \(f\) a probability density function? Justify your answer.

A sawmill wants to assess the performance of a debarker. They determine that length of time between machine failures is exponentially distributed with probability density function

where \(t\) is measured in hours.

1. Determine the probability that the debarker breaks down between \(t=200\) and \(t=600\) hours.

   Answer
   0.318
   Solution
   We compute:

   \begin{equation*} P(200 \leq t \leq 600) = \int_{200}^{600} 0.005 e^{-0.005t}\,dt = -e^{-0.005t}\big\vert_{200}^{600} \approx 0.318. \end{equation*}

2. Determine the probability that the machine breaks down after \(t=1000\) hours.

   Answer
   0.0067
   Solution
   The probability that the machine breaks down after 1000 hours is

   \begin{equation*} P(t \geq 1000) = \lim_{u\to\infty} \int_{1000}^u 0.005e^{-0.005t}\,dt = \lim_{u\to\infty} \left[-e^{-0.005t}\right]_{1000}^u = e^{-5} \approx 0.0067. \end{equation*}

For each of the given probability density functions \(f(x)\text{,}\) determine (i) the mean, (ii) the variance, and (iii) the standard deviation.

1. \(f(x)=\dfrac{3}{215}x^2, \ x \in[1,6]\)

   Answer
   (i) 4.52 (ii) 17.18 (ii) 4.14
   Solution
   We find the mean \(\mu\text{,}\) variance \(V\) and standard deviation \(\sigma\) of the probability density function

   \begin{equation*} f(x)=\dfrac{3}{152} x^2 \text{ for } x\in [1,6]. \end{equation*}

   1. \begin{equation*} \begin{aligned} \mu \amp= \int_{-\infty}^{\infty} xf(x)\,dx = \int_1^6 \frac{3}{215} x^3 \,dx\\ \amp= \frac{777}{172} \approx 4.52.\end{aligned} \end{equation*}
   2. \begin{equation*} \begin{aligned} V \amp= \int_{-\infty}^{\infty} x^2f(x)\,dx - \mu^2 = \int_1^6 \frac{3}{215}x^4\,dx - \frac{777}{172}\\ \amp= \frac{2955}{172} \approx 17.18.\end{aligned} \end{equation*}

   3. \(\sigma = \sqrt{V} \approx 4.14\)

2. \(f(x)=\dfrac{1}{36}(x-1)(7-x), \ x \in[1,7]\)

   Answer
   (i) 4 (ii) 13.8 (iii) 3.7
   Solution
   We find the mean \(\mu\text{,}\) variance \(V\) and standard deviation \(\sigma\) of the probability density function

   \begin{equation*} f(x)=\dfrac{1}{36} (x-1)(7-x) \text{ for } x\in [1,7]. \end{equation*}

   1. \begin{equation*} \begin{aligned} \mu \amp= \int_{-\infty}^{\infty} xf(x)\,dx \\ \amp= \int_1^7 \frac{1}{36} x(x-1)(7-x) \,dx = 4. \end{aligned} \end{equation*}
   2. \begin{equation*} \begin{aligned} V \amp= \int_{-\infty}^{\infty} x^2f(x)\,dx - \mu^2 = \int_1^7 \frac{1}{36} x^2(x-1)(7-x) \,dx - 4 = \\ \amp= \frac{89}{5} \approx 13.8.\end{aligned} \end{equation*}

   3. \(\sigma = \sqrt{V} \approx 3.7\)

3. \(f(x)=\dfrac{24}{x^4}, \ x \in[2,\infty)\)

   Answer

   i.3, ii.3, iii. \(\sqrt{3}\)

   Solution

   We find the mean \(\mu\text{,}\) variance \(V\) and standard deviation \(\sigma\) of the probability density function

   \begin{equation*} f(x)=\dfrac{24}{x^4} \text{ for } x\in [2,\infty)\text{.} \end{equation*}

   1. \(\begin{aligned}\mu \amp = \int_{-\infty}^{\infty}xf(x)\,dx = 24\int_2^{\infty}\frac{1}{x^3}\,dx\\ \amp = -\frac{24}{2} \lim_{a\to\infty} \frac{1}{x^2} \bigg\vert_2^a = 12 \left(\frac{1}{4} - \lim_{a\to\infty}\frac{1}{a^2}\right) = \frac{12}{4} = 3 \end{aligned}\)

   2. We first calculate

      \begin{equation*} \begin{split} \int_{-\infty}^{\infty} x^2f(x)\,dx \amp= 24 \int_2^{\infty} \frac{1}{x^2}\,dx \\ \amp= -24 \lim_{a\to\infty}\frac{1}{x}\bigg\vert_2^a \\ \amp= 24\left(\frac{1}{2} - \lim_{a\to\infty}\frac{1}{a}\right) = \frac{24}{2} = 12\end{split}\text{.} \end{equation*}

      Therefore, using our computed value \(\mu\text{,}\) the variance of the distribution is

      \begin{equation*} V(x) = \int_{-\infty}^{\infty}x^2f(x)\,dx - \mu^2 = 12 - 9 = 3\text{.} \end{equation*}

   3. \(\sigma = \sqrt{V(x)} = \sqrt{3}\)

4. \(f(x)=\dfrac{1}{5}e^{-x/5}, \ x \in [0,\infty)\)

   Answer
   Solution
   We find the mean \(\mu\text{,}\) variance \(V\) and standard deviation \(\sigma\) of the probability density function

   \begin{equation*} f(x)=\dfrac{1}{5} e^{-x/5} \text{ for } x\in [0,\infty). \end{equation*}

   1. \begin{equation*} \begin{aligned} \mu \amp= \int_{-\infty}^{\infty} xf(x)\,dx = \int_0^{\infty} \dfrac{1}{5} xe^{-x/5}\,dx \\ \amp= \lim_{a\to\infty} \left[-e^{-x/5}(x+5)\right]_0^a = 5\end{aligned} \end{equation*}
   2. \begin{equation*} \begin{aligned} V \amp= \int_{-\infty}^{\infty} x^2f(x)\,dx - \mu^2 = \int_0^{\infty} \dfrac{1}{5} x^2e^{-x/5}\,dx - 5\\ \amp= 50 - 5 = 45. \end{aligned} \end{equation*}

   3. \(\sigma = \sqrt{V} \approx 6.7\)

Suppose the probability density function which describes the number of leaves on a certain plant is given by

1. Determine the probability that a plant has exactly one leaf.

   Answer
   0.2
   Solution

   Let \(X\) be the discrete random variable which indicates the number of leaves on a plant. The probability that a plant has exactly one leaf is

   \begin{equation*} P(X=1) = f_1 = \frac{1}{20}(5-1) =\frac{1}{5} = 0.2\text{.} \end{equation*}

2. Determine the probability that a plant has less than 3 leaves.

   Answer
   0.5
   Solution

   Let \(X\) be the discrete random variable which indicates the number of leaves on a plant. The probability that a plant has less than 3 leaves is

   \begin{equation*} P(X=0)+P(X=1)+P(X=2) = f_0+f_1+f_2 = \frac{1}{20} \left(0+4+6\right)=\frac{1}{2} = 0.5\text{.} \end{equation*}

A city planner wishes to improve traffic on a busy bridge. He determines that the length of time it takes a car to cross the bridge is a continuous random variable with probability density function

where \(t\) is measured in minutes. How long is a randomly chosen car expected to take to cross the bridge? \鈥媌egin{ans} 2 minutes \end{ans}

The amount of chicken (in kg) demanded weekly at a popular restaurant is a continuous random variable with probability distribution function

What is the expected weekly demand for chicken? Answer

Let \(X\) and \(Y\) be a pair of continuous random variables with the given joint probability density function. Are \(X\) and \(Y\) independent?

1. \(f_{XY}(x,y) = \begin{cases}6xy \amp 0 \leq x \leq 1, 0 \leq y \leq \sqrt{x}\\ 0\amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

   Answer
   Not independent.
   Solution

   We compute the marginal probability density functions:

   \begin{equation*} f_X(x) = \int_0^{\sqrt{x}} 6xy \,dy = 3xy^2 \big\vert_{y=0}^{\sqrt{x}} = 3x^2\text{,} \end{equation*}

   and

   \begin{equation*} f_Y(y) = \int_0^1 6xy \,dx = 3yx^2 \big\vert_0^1 = 3y\text{.} \end{equation*}

   Since

   \begin{equation*} f_X(x)\cdot f_Y(y) = (3x^2)(3y) = 9x^2y \neq f(x,y)=6xy\text{,} \end{equation*}

   we find that \(X\) and \(Y\) are not independent.

2. \(f_{XY}(x,y) = \begin{cases}6e^{-(2x+3y)} \amp x,y \geq 0 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

   Answer
   Independent.
   Solution
   We compute the marginal probability density functions:

   \begin{equation*} f_X(x) = \int_0^{\infty} 6e^{-(2x+3y)}\,dy = 2 \lim_{a\to\infty} -e^{-(2x+3y)} \big\vert_0^a = 2e^{-2x}, \end{equation*}

   and

   \begin{equation*} f_Y(y) = \int_0^{\infty} 6e^{-(2x+3y)}\,dx = 3 \lim_{a\to\infty} -e^{-(2x+3y)} \big\vert_0^a = 3e^{-3y}. \end{equation*}

   Since

   \begin{equation*} f_X(x)\cdot f_Y(y) = \bigl( 2e^{-2x}\bigr)\bigl( 3e^{-3y}\bigr) = 6e^{-(2x+3y)} = f(x,y), \end{equation*}

   for \(x, y \geq 0\) we find that \(X\) and \(Y\) are independent.

3. \(f_{XY}(x,y) = \begin{cases}10x^2y \amp 0 \leq y \leq x \leq 1 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

   Answer
   Not independent.
   Solution
   Note that the region of integration is the triangle with vertices at \((0,0)\text{,}\) \((1,1)\) and \((1,0)\text{.}\) We compute the marginal probability density functions:

   \begin{equation*} f_X(x) = \int_0^x 10x^2y\,dy = 5x^4, \quad \text{for } 0 \leq x \leq 1, \end{equation*}

   and

   \begin{equation*} f_Y(y) =\int_y^1 10x^2y\,dx = \frac{10}{3} (y-y^4), \quad \text{for } 0 \leq y\leq 1. \end{equation*}

   Since

   \begin{equation*} f_X(x)\cdot f_Y(y) = \bigl( 5x^4 \bigr)\bigl(\frac{10}{3} (y-y^4) \bigr) \neq f(x,y) = 10x^2 y \end{equation*}

   on the triangular domain, we find that \(X\) and \(Y\) are not independent.

4. \(f_{XY}(x,y) = \begin{cases}2 \amp 0 \leq x\leq y \leq 1 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

   Answer
   Not independent.
   Solution
   Note that the region of integration is the triangle with vertices at \((0,0)\text{,}\) \((1,1)\) and \((0,1)\text{.}\) We compute the marginal probability density functions:

   \begin{equation*} f_X(x) = \int_x^1 2\,dy = 2(1-x) \quad \text{ for } 0 \leq x \leq 1, \end{equation*}

   and

   \begin{equation*} f_Y(y) = \int_0^y 2\,dx = 2y \quad \text{ for } 0 \leq y \leq 1. \end{equation*}

   Since

   \begin{equation*} f_X(x)\cdot f_Y(y) = \bigl(2(1-x) \bigr)\bigl(2y \bigr) \neq f(x,y)=2 \end{equation*}

   on the domain in question, we find that $X$ and $Y$ are independent.

Given the following probability density function \(f(x,y)\) for a pair of continuous random variables \(X\) and \(Y\text{,}\) compute (i) the expected values \(E(X)\) and \(E(Y)\text{,}\) (ii) the variance \(V(X)\) and \(V(Y)\text{,}\) and (iii) the covariance \(Cov(X,Y)\text{.}\)

1. \(f_{XY}(x,y) = \begin{cases}6xy \amp 0 \leq x \leq 1, 0 \leq y \leq \sqrt{x}\\ 0\amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

   Answer
   i.3/4,4/7; ii.3/80,19/392; iii.1/63
   Solution

   We consider the joint probability density function

   \begin{equation*} f_{XY}(x,y) = \begin{cases}6xy \amp 0 \leq x \leq 1, 0 \leq y \leq \sqrt{x} \\ 0 \amp \text{ all other \(x\) and \(y\). } \end{cases} \end{equation*}

   1. The expected value of \(X\) is

      \begin{equation*} E(X) = \int_0^1 \int_0^{\sqrt{x}} 6x^2y\,dy\,dx = \int_0^1 3x^3 \,dx = \frac{3}{4}x^4 \bigg\vert_0^1 = \frac{3}{4}\text{,} \end{equation*}

      and the exepected value of \(Y\) is

      \begin{equation*} E(Y) = \int_0^1\int_0^{\sqrt{x}} 6xy^2 \,dy\,dx = \int_0^1 2x^{5/2}\,dx = \frac{4}{7}x^{7/2}\bigg\vert_0^1 = \frac{4}{7}\text{.} \end{equation*}

   2. First, we compute

      \begin{equation*} \int_0^1\int_0^{\sqrt{x}} 6x^3y\,dy\,dx = \int_0^1 3x^4\,dx = \frac{3}{5}x^5\bigg\vert_0^1 = \frac{3}{5}\text{,} \end{equation*}

      and

      \begin{equation*} \int_0^1\int_0^{\sqrt{x}} 6xy^3\,dy\,dx = \int_0^1 \frac{3}{2}x^3 \,dx = \frac{3}{8}x^4\bigg\vert_0^1 = \frac{3}{8}\text{.} \end{equation*}

      Using the calculated values for \(E(X)\) and \(E(Y)\text{,}\) we find the variance of \(X\) to be

      \begin{equation*} V(X) = \int_0^1\int_0^{\sqrt{x}} 6x^3y\,dy\,dx - E(X)^2 = \frac{3}{5} - \frac{9}{16} = \frac{3}{80}\text{,} \end{equation*}

      and the variance of \(Y\) to be

      \begin{equation*} V(Y)=\int_0^1\int_0^{\sqrt{x}} 6xy^3\,dy\,dx-E(Y)^2 = \frac{3}{8}-\frac{16}{49} = \frac{19}{392}\text{.} \end{equation*}

   3. First calculate

      \begin{equation*} \int_0^1\int_0^{\sqrt{x}} 6x^2y^2\,dy\,dx = \int_0^1 2x^{7/2}\,dx = \frac{4}{9}x^{9/2}\bigg\vert_0^1 = \frac{4}{9}\text{.} \end{equation*}

      Then, again using our computed values for \(E(X)\) and \(E(Y)\text{,}\) the covariance of \(X\) and \(Y\) is

      \begin{equation*} Cov(X,Y) = \int_0^1\int_0^{\sqrt{x}} 6x^2y^2\,dy\,dx - E(X)E(Y)=\frac{4}{9}-\frac{3}{4}\cdot\frac{4}{7} =\frac{1}{63} \approx 0.016\text{.} \end{equation*}

      Thus, \(X\) and \(Y\) are slightly positively correlated.

2. \(f_{XY}(x,y) = \begin{cases}6e^{-(2x+3y)} \amp x,y \geq 0 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

3. \(f_{XY}(x,y) = \begin{cases}10x^2y \amp 0 \leq y \leq x \leq 1 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)

4. \(f_{XY}(x,y) = \begin{cases}2 \amp 0 \leq x\leq y \leq 1 \\ 0 \amp \text{ all other \(x\) and \(y\) } . \end{cases}\)