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《计算机视觉》习题解答（一）第2章_计算机视觉基础课第二章答案

作者：2023面试高手 | 2024-05-30 22:29:04

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计算机视觉基础课第二章答案

文章目录

- 习题解答
- - - 2.1
    - 2.2
    - 2.3
    - 2.4
    - 2.5
    - 2.6
    - 2.7
    - 2.8
    - 2.9
    - 2.10

习题解答

$\color{red}{答案仅供参考}$

2.1

$统计一周的下雨量，x\ 的值为周一到周日，y\ 的值为降雨量$

2.2

$仅考虑变量\ w\ 和\ y$

\begin{aligned} P r (v, w, x, y, z) & a m p; = P r (v, x, z, w | y) P r (y) \\ a m p; = P r (v, x, z | w, y) P r (w | y) P r (y) \\ a m p; = P r (v, x, z | w, y) P r (w, y) \end{aligned}

$\begin{aligned} Pr(v,w,x,y,z) & =Pr(v,x,z,w|y)Pr(y) \\ & = Pr(v,x,z|w,y)Pr(w|y)Pr(y) \\ & =Pr(v,x,z|w,y)Pr(w,y) \\ \end{aligned}$

P r (v, w, x, y, z) = P r (v, x, z, w ∣ y) P r (y) = P r (v, x, z ∣ w, y) P r (w ∣ y) P r (y) = P r (v, x, z ∣ w, y) P r (w, y)

整 理 得 到

Pr(w,y)=\frac{Pr(v,w,x,y,z)}{Pr(v,x,z|w,y)}

同 理 可 得

Pr(v)=\frac{Pr(v,w,x,y,z)}{Pr(w,x,y,z|v)}

2.3

\begin{aligned} P r (w, x, y, z) & a m p; = P r (z | w, x, y) P r (w, x, y) \\ a m p; = P r (z | w, x, y) P r (w | x, y) P r (x, y) \end{aligned}

$\begin{aligned} Pr(w,x,y,z) & =Pr(z|w,x,y) Pr(w,x,y) \\ & =Pr(z|w,x,y)Pr(w|x,y)Pr(x,y) \end{aligned}$

P r (w, x, y, z) = P r (z ∣ w, x, y) P r (w, x, y) = P r (z ∣ w, x, y) P r (w ∣ x, y) P r (x, y)

得 证

2.4

\begin{aligned} P r (c = 2 | h = 1) & a m p; = \frac{P r (h = 1, c = 2)}{P r (h = 1)} \\ a m p; = \frac{P r (h = 1, c = 2)}{P r (h = 1, c = 1) + P r (h = 1, c = 2)} \\ a m p; = \frac{P r (h = 1 | c = 2) P r (c = 2)}{P r (h = 1, c = 1) P r (c = 1) + P r (h = 1 | c = 2) P r (c = 2)} \\ a m p; = \frac{0.5 \times 0.8}{0.5 \times 0.5 + 0.5 \times 0.8} \\ a m p; = \frac{8}{13} \end{aligned}

$\begin{aligned} Pr(c=2|h=1) & =\frac{Pr(h=1,c=2)}{Pr(h=1)} \\ & =\frac{Pr(h=1,c=2)}{Pr(h=1,c=1)+Pr(h=1,c=2)} \\ & =\frac{Pr(h=1|c=2)Pr(c=2)}{Pr(h=1,c=1)Pr(c=1)+Pr(h=1|c=2)Pr(c=2)} \\ & =\frac{0.5\times0.8}{0.5\times0.5+0.5\times0.8} \\ & =\frac{8}{13} \end{aligned}$

P r (c = 2 ∣ h = 1) = \frac{P r ( h = 1 , c = 2 )}{P r ( h = 1 )} = \frac{P r ( h = 1 , c = 2 )}{P r ( h = 1 , c = 1 ) + P r ( h = 1 , c = 2 )} = \frac{P r ( h = 1 ∣ c = 2 ) P r ( c = 2 )}{P r ( h = 1 , c = 1 ) P r ( c = 1 ) + P r ( h = 1 ∣ c = 2 ) P r ( c = 2 )} = \frac{0 . 5 \times 0 . 8}{0 . 5 \times 0 . 5 + 0 . 5 \times 0 . 8} = \frac{8}{1 3}

2.5

$y\ 和\ z\ 不相互独立。考虑任意一个概率分布Pr(y,z)并且\ y\ 和\ z\ 不独立，\\那么我们可以找到一个概率分布Pr(x)，它并没有提供关于\ y\ 和\ z\ 的任何信息\\ 即\ x\ 和\ y\ 相互独立且\ x\ 和\ z\ 相互独立$

2.6

\begin{aligned} P r (x | y = y^{*}) & a m p; = \frac{P r (x, y = y^{*})}{P r (y = y^{*})} \\ a m p; = \frac{P r (x) P r (y = y^{*})}{P r (y = y^{*})} (因 为 x 和 y 相 互 独 立) \\ a m p; = P r (x) \end{aligned}

$\begin{aligned} Pr(x|y=y^*) & =\frac{Pr(x,y=y^*)}{Pr(y=y^*)} \\ & =\frac{Pr(x)Pr(y=y^*)}{Pr(y=y^*)} (因为x和y相互独立)\\ & =Pr(x) \end{aligned}$

P r (x ∣ y = y^{*}) = \frac{P r ( x , y = y ^{*} )}{P r ( y = y ^{*} )} = \frac{P r ( x ) P r ( y = y ^{*} )}{P r ( y = y ^{*} )} (因 为 x 和 y 相 互 独 立) = P r (x)

得证

2.7

$（这道题不太懂，给定的条件不是已经证明了独立性了吗？）$

\begin{aligned} P r (w, x, y, z) & a m p; = P r (w) P r (z | y) P r (y | x, w) P r (x) \\ a m p; = P r (z | y) P r (y | x, w) P r (x, w) （ 代 入 P r (x, w) = P r (x) P r (w) ） \\ a m p; = P r (z | y) P r (x, y, w) \\ 又 有 P r (w, x, y, z) & a m p; = P r (z | x, y, w) P r (x, y, w) \end{aligned}

$\begin{aligned} Pr(w,x,y,z) & =Pr(w)Pr(z|y)Pr(y|x,w)Pr(x) \\ & =Pr(z|y)Pr(y|x,w)Pr(x,w)（代入Pr(x,w)=Pr(x)Pr(w)）\\ & =Pr(z|y)Pr(x,y,w) \\ 又有\;Pr(w,x,y,z) &=Pr(z|x,y,w)Pr(x,y,w) \end{aligned}$

P r (w, x, y, z) 又 有 P r (w, x, y, z) = P r (w) P r (z ∣ y) P r (y ∣ x, w) P r (x) = P r (z ∣ y) P r (y ∣ x, w) P r (x, w) （ 代 入 P r (x, w) = P r (x) P r (w) ） = P r (z ∣ y) P r (x, y, w) = P r (z ∣ x, y, w) P r (x, y, w)

所 以 ， P r (z ∣ y) = P r (z ∣ x, y, w) ， 又

\begin{aligned} P r (w, x, y, z) & a m p; = P r (z | x, y, w) P r (y | x, w) P r (x | w) P r (w) \\ a m p; = P r (z | y) P r (y | x, w) P r (x | w) P r (w) \\ a m p; \\ P r (w, x, y, z) & a m p; = P r (w) P r (z | y) P r (y | x, w) P r (x) \end{aligned}

对 比 两 式 ， 可 以 得 到 P r (x ∣ w) = P r (x) ， 同 理 可 得 P r (w ∣ x) = P r (w)

$所以， x 和 w 相互独立$

2.8

$（为啥概率总和不为 1 ）$

$设第一次骰子朝上的值为\bm X，第二次为\bm Y$

\begin{aligned} E [\bm X] = & a m p; 1 \times 1 / 12 + 2 \times 1 / 12 + 3 \times 1 / 12 \\ a m p; + 4 \times 1 / 12 + 5 \times 1 / 6 + 6 \times 1 / 12 \\ = & a m p; 13 / 6 \end{aligned}

$\begin{aligned} E[\bm X] =&1\times 1/12+2\times 1/12+3\times 1/12 \\ &+4\times 1/12+5\times 1/6+6\times 1/12 \\ =&13/6 \end{aligned}$

E [X] = = 1 \times 1 / 12 + 2 \times 1 / 12 + 3 \times 1 / 12 + 4 \times 1 / 12 + 5 \times 1 / 6 + 6 \times 1 / 12 13 / 6

计 算 两 次 投 掷 可 以 列 表 计 算

\begin{aligned} E [\bm X + \bm Y] = & a m p; 2 \times 1 / 144 + 3 \times 2 / 144 + 4 \times 3 / 144 \\ a m p; + 5 \times 4 / 144 + 6 \times 7 / 144 + 7 \times 8 / 144 \\ a m p; + 8 \times 7 / 144 + 9 \times 6 / 144 + 10 \times 6 / 144 \\ a m p; + 11 \times 4 / 144 + 12 \times 1 / 144 \\ = & a m p; 91 / 36 \end{aligned}

2.9

$首先我们假定随机变量 x 为连续值$

\begin{aligned} E [k] & a m p; = \int k P r (x) d x \\ a m p; = k \int P r (x) d x \\ a m p; = k \end{aligned}

$\begin{aligned} E[k] & =\int kPr(x)\text dx \\ & =k\int Pr(x)\text dx \\ & =k \end{aligned}$

E [k] = \int k P r (x) d x = k \int P r (x) d x = k

性 质 1 得 证

\begin{aligned} E [k f [x]] & a m p; = \int k f [x] P r (x) d x \\ a m p; = k \int f [x] P r (x) d x \\ a m p; = k E [f [x]] \end{aligned}

性 质 2 得 证

\begin{aligned} E [f [x] + g [x]] & a m p; = \int (f [x] + g [x]) P r (x) d x \\ a m p; = \int f [x] P r (x) d x + \int g [x] P r (x) d x \\ a m p; = E [f [x]] + E [g [x]] \end{aligned}

性 质 3 得 证

\begin{aligned} E [f [x] g [y]] & a m p; = \iint f [x] g [y] P r (x, y) d x d y \\ a m p; = \iint f [x] g [y] P r (x) P r (y) d x d y \\ a m p; = \iint f [x] P r (x) d x \cdot g [y] P r (y) d y \\ a m p; = \int f [x] P r (x) d x \int g [y] P r (y) d y \\ a m p; = E [f [x]] E [g [y]] \end{aligned}

性 质 4 得 证

2.10

\begin{aligned} E [(x - μ)^{2}] & a m p; = E [x^{2} - 2 μ x + μ^{2}] \\ a m p; = E [x^{2}] + E [- 2 μ x] + E [μ^{2}] \\ a m p; = E [x^{2}] - 2 μ E [x] + μ^{2} \\ a m p; = E [x^{2}] - 2 E [x] E [x] + E [x] E [x] \\ a m p; = E [x^{2}] - E [x] E [x] \end{aligned}

$\begin{aligned} E[(x-\mu)^2] & =E[x^2-2\mu x+\mu^2]\\ & =E[x^2]+E[-2\mu x]+E[\mu^2]\\ & =E[x^2]-2\mu E[x]+\mu^2\\ & =E[x^2]-2E[x]E[x]+E[x]E[x]\\ & =E[x^2]-E[x]E[x] \end{aligned}$

E [(x - μ)^{2}] = E [x^{2} - 2 μ x + μ^{2}] = E [x^{2}] + E [- 2 μ x] + E [μ^{2}] = E [x^{2}] - 2 μ E [x] + μ^{2} = E [x^{2}] - 2 E [x] E [x] + E [x] E [x] = E [x^{2}] - E [x] E [x]

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