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W07 - HW Solutions

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Joint Distributions

01 - Finish a PMF table - Strange families

Suppose that 15 percent of the families in a strange community have no children, 20 percent have 1 child, 35 percent have 2 children, and 30 percent have 3 children. Assume the odds of a child being a boy or a girl are equal.

If a family is chosen at random from this community, then , the number of boys, and , the number of girls, in this family will have the joint PMF partially shown in Table 6.2:

  • (a) Complete the table by finding the missing entries.
  • (b) What is the probability that ” or is “?

Solution (a)

  1. & Fill the cells using the respective column sum or row sum.
    • We have , so
    • : We have , so
    • : We have , so
    • : We have , so

(b)

  1. & Add up the probabilities in which either or .
  2. && Alternatively, you could use the inclusion-exclusion principle using the marginal sums.

02 - PMF calculations from a table

Suppose the joint PMF of and has values given in this table:

  • (a) Find .
  • (b) Find the marginal PMF of .
  • (c) Find the PMF of the random variable .
  • (d) Find and .

Solution (a)

  1. & We know that . Use this to find .

(b)

  1. & The marginal PMF of is given by adding up the rows corresponding to each possible value.

(c)

  1. & Define the possible values of .
  • Since and , we have that .

  1. && Define PMF of .

    • Go through each possible value of and see when it occurs.

  2. & Substitute values in for each probability.

(d)

  1. & Find
  2. & Find

03 - Marginals from joint PMF

Suppose the discrete joint PMF of and is given by: Compute the marginal PMFs and .

Solution

  1. Compute by summing over .
  2. Compute by summing over .

04 - Joint CDF on box events: All four corners

Consider the following formula: Prove this formula. Hint: Do these steps along the way:

  • Draw these events in the -plane: Misplaced &A &= \{X\leq x_{1}, y_{1} < Y \leq y_{2}\} \\ B &= \{x_{1}< X \leq x_{2}, Y \leq y_{1}\} \\ C &= \{x_{1} < X \leq x_{2}, y_{1} < Y \leq y_{2}\} \end{align*}$$
  • Draw the event . Write the probability of this event in terms of .

Solution

  1. & Write the event in terms of .
  2. & Write the terms , , and in terms of . Misplaced &P[A] &= F_{X, Y}(x_{1}, y_{2}) - F_{X, Y}(x_{1}, y_{1}) \\ P[B] &= F_{X, Y}(x_{2}, y_{1}) - F_{X, Y}(x_{1}, y_{1}) \\ P[C] &= P[A \cup B \cup C] - \left(P[A] + P[B]\right) \end{align*} $$
  3. && Simplify expression for . Misplaced &P[C] &= F_{X, Y}(x_{2}, y_{2}) - F_{X, Y}(x_{1}, y_{1}) - F_{X, Y}(x_{1}, y_{2}) - F_{X, Y}(x_{2}, y_{1}) + 2F_{X, Y}(x_{1}, y_{1}) \\ &= F_{X, Y}(x_{2}, y_{2}) - F_{X, Y}(x_{2}, y_{1}) - F_{X, Y}(x_{1}, y_{2}) + F_{X, Y}(x_{1}, y_{2})\end{align*}$$

05 - Marginals from PDF

Suppose and have joint PDF given by

  • (a) Find the marginal PDFs for and .
  • (b) Find .

Solution

(a)

  1. & Find the marginal PDF for by integrating the joint PDF with respect to .
  2. & Find the marginal PDF for by integrating the joint PDF with respect to .

(b)

  1. && Note that the region of interest is the one above the line . Integrate the PDF over this region.

06 - Random point in a triangle

Consider a joint distribution whose PDF is constant inside the triangle with , , and , and zero outside. Suppose a point is chosen at random according to this distribution.

  • (a) Find the joint PDF .
  • (b) Find the marginal PDFs for and .
  • (c) Are and independent?

Solution

(a)

  1. & Find the area of the triangle, and find a formula for the PDF.
    • The area of the triangle is . Therefore the PDF is
    Misplaced &\frac{1}{1/2} = 2 & 0 \leq x \leq 1, 0 \leq y \leq 1 - x \\ 0 & \text{otherwise} \end{cases}$$

(b)

  1. & Integrate with respect to to find the marginal PDF for .
Misplaced &2(1-x) & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ 2. & Integrate with respect to $x$ to find the marginal PDF for $Y$. $$f_{Y}(y) = \int_{0}^{1-y}2dx = \left.2x\right|_{0}^{1-y} = 2(1-y)$$ $$f_{Y}(y) = \begin{cases} 2(1- y) & 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ **(c)** 1. & Compute the product $f_{X}(x)f_{Y}(y)$ and compare to $f_{X, Y}(x, y)$. $$2 \stackrel{?}{=} 4(1-x)(1-y) $$ 2. & Consider the case wherein $x = 1, y = 0$ $$4(1-1)(1-0) = 0 \neq 2$$ - Therefore, $X$ and $Y$ are not independent. #### 07 - Factorizing the density Consider two joint density functions for $X$ and $Y$: $$\begin{align*} f_{1}(x, y) &= 6e^{-2x}e^{-3y}, & x, y > 0, \\ f_{2}(x, y) &= 2yxe^{x^{2}}, & x, y \in[0, 1], x + y \in [0, 1]. \end{align*}$$ (Assume the densities are zero outside the given domain). Supposing $f_{1}$ is the joint density, are $X$ and $Y$ independent? Why or why not? Suppose $f_{2}$ is the joint density, are $X$ and $Y$ independent? Why or why not? **Solution** 1. & Compute the marginal distribution of $X$ by integrating $f_{1}$ with respect to $y$. $$f_{X}(x) = \int_{0}^{\infty}6e^{-2x}e^{-3y}dy = e^{-2x}\lim_{b \to \infty}\left[-2e^{-3y}\right]_{0}^{b} = 2e^{-2x}, x > 0$$ 2. & Compute the marginal distribution of $Y$ by integrating $f_{1}$ with respect to $x$. $$f_{Y}(y) = \int_{0}^{\infty}6e^{-2x}e^{-3y}dx = 3e^{-3y}, x > 0$$ 3. & Determine independence by multiplying the marginal pdfs. $$f_{X}(x)f_{Y}(y) = 6e^{-2x}e^{-3y} = f_{1}$$ - Since the product of the marginal PDFs equals the joint PDF, we conclude that $X$ and $Y$ are independent. 4. & Compute the marginal distribution of $X$ by integrating $f_{2}$ with respect to $y$. $$f_{X}(x) = \int_{0}^{1-x}2yxe^{x^{2}}dy = xe^{x^{2}}\left.\left[y^{2}\right]\right|_{0}^{1-x} = xe^{x^{2}}(1-x)^{2} $$ 5. & Compute the marginal distribution of $Y$ by integrating $f_{2}$ with respect to $x$. $$f_{Y}(y) = \int_{0}^{1-y}2yxe^{x^{2}}dx = y\left.\left[e^{x^{2}}\right]\right|_{0}^{1-y} = y\left(e^{(1-y)^{2}}-1\right)$$ 6. & Determine independence by multiplying the marginal pdfs. - $f_{X}(x)f_{Y}(y) = yxe^{x^{2}}(1-x)^{2}\left(e^{(1-y)^{2}}-1\right) \neq f_{2}$ - Therefore, $X$ and $Y$ are *not* independent. #### 08 - Composite PDF from joint PDF The joint density of random variables $X$ and $Y$ is given by $$f_{X, Y}(x, y) = \begin{cases} e^{-x-y} & x, y > 0 \\ 0 & \text{otherwise} \end{cases}$$ Compute the PDF of $X/Y$. **Solution** 1. && Define $Z = X/Y$ and find the CDF of $Z$. $$\begin{align*}P[Z \leq z] &= P[X/Y \leq z] = P[X \leq Yz] \\ &= \int_{0}^{\infty}\int_{0}^{yz}e^{-x}e^{-y}dxdy \\&= -\int_{0}^{\infty}e^{-y}\left.\left[e^{-x}\right]\right|_{0}^{yz}dy \\&= 1-\int_{0}^{\infty}e^{-y(z + 1)dy} \\&= 1 - \frac{1}{z + 1}, \quad z > 0\end{align*}$$ 2. & Differentiate to find $f_{Z}(z)$. $$f_{Z}(z) = \frac{1}{(z + 1)^{2}}, \quad z > 0$$