Course Landing Page

W11 - HW Solutions

7 min read

Conditional distribution

01 - Conditional density from joint density

Suppose that and have joint probability density given by: (a) Compute , for . (b) Compute .

Solution

(a)

  1. Write out formula for conditional distribution.
  2. Compute marginal distribution for .
  3. Find conditional density function. Misplaced &\frac{6x(2-x-y)}{4-3y} & x, y \in [0, 1] \\ 0 & \text{otherwise} \end{cases}$$

(b) 4. Integrate to find the desired probability.

Misplaced &P\left[X > \frac{1}{2} \mid Y = y\right] &= \int_{\frac{1}{2}}^{1}\frac{6x(2-x-y)}{4-3y}dx \\ &= \frac{6}{4-3y}\int_{\frac{1}{2}}^{1}2x-x^{2}-xydx \\ &= \frac{6}{4-3y}\left.\left[x^{2}-\frac{x^{3}}{3}-\frac{x^{2}y}{2}\right]\right|_{\frac{1}{2}}^{1} \\ &= \frac{11-9y}{4(4-3y)} \end{align*}$$ 5. State final answer. $$P\left[X > \frac{1}{2} \mid Y = y\right] = \begin{cases} \frac{11-9y}{4(4-3y)} & y \in [0, 1] \\ 0 & \text{otherwise} \end{cases}$$ #### 02 - From conditional to joint, and back again Suppose we have the following data about random variables $X$ and $Y$: $$\begin{align*} f_{X}(x) &= \begin{cases} 3x^{2} & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases} \\ f_{Y \mid X}(y \mid x) &= \begin{cases} 2y/x^{2} & 0 \leq y \leq x \\ 0 & \text{otherwise} \end{cases} \end{align*}$$ (a) Find the joint distribution $f_{X, Y}(x, y)$. (b) Find $f_{X \mid Y}(x \mid y)$. **Solution** **(a)** 1. State formula $$f_{Y \mid X}(y\mid x)= \frac{f_{X, Y}(x, y)}{f_{X}(x)} \implies f_{X, Y}(x, y) = f_{Y \mid X}(y \mid x)f_{X}(x)$$ 2. Find joint distribution. $$f_{X, Y}(x, y) = \begin{cases} 3x^{2} \frac{2y}{x^{2}} = 6y & 0 \leq y \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ **(b)** 3. State formula $$f_{X\mid Y}(x \mid y) = \frac{f_{X, Y}(x, y)}{f_{Y}(y)}$$ 4. Find marginal distribution of $Y$. $$f_{Y}(y) = \int_{y}^{1}6ydx = \left.6yx\right|_{y}^{1} = 6y - 6y^{2} = 6y(1-y)$$ 5. Find conditional distribution. $$f_{X \mid Y}(x \mid y) = \frac{f_{X, Y}(x, y)}{f_{Y}(y)} = \begin{cases} \frac{1}{1-y} & 0 \leq y \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ ## Conditional expectation #### 03 - Conditional distribution and expectation from joint PMF Suppose that $X$ and $Y$ have the following joint PMF: $$P_{X, Y}(k, \ell) = \begin{cases} c & k = 1, 2, 3, 4; \quad \ell = 1,\dots, k \\ 0 & \text{otherwise}\end{cases}$$ Notice that the possibilities for $\ell$ depend on the choice of $k$. First, show that $c = \frac{1}{10}$ must be true. Then, compute (a) $P_{X}$ (b) $P_{Y \mid X}$ (c) $E[Y \mid X = 4]$ (d) $E[Y \mid X]$ **Solution** 1. Find possible combinations of $k$ and $\ell$. - For a given value of $k$, there are $k$ possible options for $\ell$. - Thus, in total, there are $1 + 2 + 3 + 4 = 10$ possible combinations. - Since $10 \cdot c = 1$, we have that $c = \frac{1}{10}$. **(a)** 2. We have that $$\begin{align*} P_{X}(1) &= \frac{1}{10} \\ P_{X}(2) &= \frac{2}{10} \\ P_{X}(3) &= \frac{3}{10} \\ P_{X}(4) &= \frac{4}{10} \end{align*}$$ Thus, $$P_{X}(k) = \begin{cases} \frac{k}{10} & k = 1, 2, 3, 4\\ 0 & \text{otherwise} \end{cases}$$ **(b)** 1. State formula $$P_{Y \mid X}(\ell \mid k) = \frac{P_{X, Y}(k, \ell)}{P_{X}(k)} = \frac{c}{kc} = \frac{1}{k}$$ 2. State final answer $$P_{Y \mid X}(\ell \mid k) = \begin{cases} \frac{1}{k} & \ell = 1, \dots, k \\ 0 & \text{otherwise} \end{cases}$$ **(c)** 3. Compute expectation: $$E[Y \mid X = 4] = \sum_{\ell = 1}^{4}\frac{\ell}{4} = \frac{5}{2}$$ **(d)** 4. Find formula for expectation. $$E[Y \mid X] = \sum_{\ell = 1}^{x} \frac{\ell}{x} = \frac{1}{x}\sum_{\ell = 1}^{x}\ell = \frac{1}{x}\frac{x(x + 1)}{2} = \frac{x+ 1}{2}$$ #### 04 - Conditional distribution and expectation from joint PDF Suppose that $X$ and $Y$ have the following joint PDF: $$f_{X, Y}(x, y) = \begin{cases} cxy & 0 < y < 1, 0 < x < y \\ 0 & \text{otherwise} \end{cases}$$ Notice that the range of possibilities for $x$ depends on the choice of $y$. First, show that $c = 8$ must be true. Then compute: (a) $f_{X}$ (b) $f_{Y \mid X}$ (c) $E[Y \mid X = 0.5]$ (d) $E[Y \mid X]$ **Solution** 1. Integrate joint PDF and solve for $c$. $$\begin{align*} \int_{0}^{1}\int_{0}^{y}cxydxdy &= 1 \\ \int_{0}^{1}\frac{cy^{3}}{2}dy &= 1\\ \frac{c}{8}&=1 \implies c = 8 \end{align*}$$ **(a)** 2. Integrate joint PDF with respect to $y$ to obtain $f_{X}$ $$f_{X}(x) = \int_{x}^{1}8xydy = 8x\left.\left[\frac{y^{2}}{2}\right]\right|_{x}^{1} = 4x\left(1-x^{2}\right)$$ **(b)** 1. Use formula for conditional distribution. $$f_{Y \mid X}(y \mid x) = \frac{f_{x, y}(x, y)}{f_{X}(x)} = \frac{8xy}{4x\left(1-x^{2}\right)} = \frac{2y}{1- x^{2}}, \qquad x < y$$ **(c)** 1. Plug in $x = 0.5$ into the formula found in part (b) $$f_{Y \mid X}(y\mid 0.5) = \frac{2y}{1-0.5^{2}} = \frac{8y}{3}$$ 2. Integrate to find expectation. $$E[Y \mid X = 0.5] = \int_{0.5}^{1}\frac{8y^{2}}{3}dy = \frac{8}{3}\left.\left[\frac{y^{3}}{3}\right]\right|_{0.5}^{1} = \frac{7}{9}$$ **(d)** Integrate formula in part (b) with respect to $y$. $$E[Y \mid X = x] = \int_{x}^{1}\frac{2y^{2}}{1-x^{2}}dy =\frac{2}{1-x^{2}}\left.\left[\frac{y^{3}}{3}\right]\right|_{x}^{1} = \frac{2\left(1-x^{3}\right)}{3\left(1-x^{2}\right)}$$ #### 05 - "Plug In" Expectation Identity Suppose $h(x, y)$ is a function, and $X$ and $Y$ are two random variables. Verify this formula in the *continuous case*, using the definitions: $$E\left[h(X, Y)\mid Y = y\right] = E\left[h(X, y)\mid Y = y\right]$$ Using that formula, prove this formula: $$E\left[a(Y)b(X)\mid Y\right] = a(Y)E\left[b(X)\mid Y\right]$$ for two functions $a(y)$ and $b(x)$ and random variables $X$ and $Y$. Notice that here the expectations are viewed as random variables. **Solution** 1. State formula for expectation given $Y= y$. $$E\left[h(X, Y)\mid Y = y\right] = \int_{-\infty}^{\infty}h(x, y)f_{X \mid Y}(x \mid y)dx$$ 2. Note that $Y =y$ is fixed inside the conditional expectation. - Treat $h(X, Y)$ as $h(X, y)$ inside integral. $$E\left[h(X, Y)\mid Y = y\right] = \int_{-\infty}^{\infty}h(x, y)f_{X \mid Y}(x \mid y)dx = E\left[h(X, y)\mid Y = y\right]$$ 3. Prove the second formula. $$E\left[a(Y)b(X)\mid Y = y\right] = \int a(y)b(x)f_{X \mid Y}(x \mid y)dx = a(y)\int b(x)f_{X \mid Y}(x \mid y)dx = a(y)E\left[b(X)\mid Y = y\right]$$ 4. State final answer. $$E\left[a(Y)b(X) \mid Y\right] = a(Y)E\left[b(X) \mid Y\right]$$ #### 06 - Iterated Expectation Identity Prove the following identity using iterated expectation along with the previous exercise: $$E[XY] = E\left[YE[X \mid Y]\right]$$ **Solution** 1. Let $a(Y) = Y$ and $b(X) = X$. $$E[XY] = E\left[E\left[b(X)a(Y)\mid Y\right]\right] = E\left[a(Y)E\left[b(X)\mid Y\right]\right] = E\left[Y E[X \mid Y]\right]$$ #### 07 - How many customers buy a cake? Let $N$ count the number of customers that visit a bakery on a random day, and assume $N \sim \text{Pois}(\lambda)$. Let $X$ count the number of customers that make a purchase. Each customer entering the bakery smells the cakes, and this produces a probability $p$ of buying a cake for that customer. The customers are independent. Find $\text{Cov}[N, X]$. Are $N$ and $X$ positively or negatively correlated? **Solution** 1. Find $E[X \mid N]$. - Note that this follows a binomial distribution with parameters $n, p$. - Thus, $E[X \mid N] = Np$ 2. Find $E[X]$ using iterated expectation. $$E[X] = E\left[E[X \mid N]\right] = E[Np] = p\lambda$$ 3. Find $E[NX]$ $$\begin{align*} E[NX] &= E\left[NE[X \mid N]\right] = E\left[N^{2}p\right] = pE\left[N^{2}\right] \\ E\left[N^{2}\right] &= \text{Var}[N] + E\left[N^{2}\right] = \lambda + \lambda^{2} \\ E[NX] &= p\left(\lambda + \lambda^{2}\right) \end{align*}$$ 4. Find covariance $$\text{Cov}[N, X] = E[NX] - E[N]E[X] = p\left(\lambda + \lambda^{2}\right) - p\lambda^{2} = p\lambda$$ 5. State final conclusions. - Since the covariance is positive, $N$ and $X$ are *positively* correlated.