本文为 I n t r o d u c t i o n Introduction Introduction t o to to P r o b a b i l i t y Probability Probability 的读书笔记
Joint PDF
- We say that two continuous random variables associated with the same experiment are jointly continuous (联合连续的) and can be described in terms of a joint PDF f X , Y f_{X,Y} fX,Y if f X , Y f_{X,Y} fX,Y is a nonnegative function that satisfies
P ( ( X , Y ) ∈ B ) = ∫ ∫ ( x , y ) ∈ B f X , Y ( x , y ) d x d y P((X,Y)\in B)=\int\int_{(x,y)\in B} f_{X,Y}(x,y)dxdy P((X,Y)∈B)=∫∫(x,y)∈BfX,Y(x,y)dxdyfor every subset B B B of the two-dimensional plane. In the particular case where B B B is a rectangle of the form B = { ( x , y ) ∣ a ≤ x ≤ b , c ≤ y ≤ d } B=\{(x,y)|a\leq x\leq b,c\leq y\leq d\} B={ (x,y)∣a≤x≤b,c≤y≤d}, we have
P ( a ≤ X ≤ b , c ≤ Y ≤ d ) = ∫ c d ∫ a b f X , Y ( x , y ) d x d y P(a\leq X\leq b,c\leq Y\leq d)=\int_c^d\int_a^bf_{X,Y}(x,y)dxdy P(a≤X≤b,c≤Y≤d)=∫cd∫abfX,Y(x,y)dxdyFurthermore, by letting B B B be the entire two-dimensional plane, we obtain the normalization property
∫ − ∞ ∞ ∫ − ∞ ∞ f X , Y ( x , y ) d x d y = 1 \int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)dxdy=1 ∫−∞∞∫−∞∞fX,Y(x,y)dxdy=1 - To interpret the joint PDF, we let δ \delta δ be a small positive number and consider the probability of a small rectangle. We have
P ( a ≤ X ≤ a + δ , c ≤ Y ≤ c + δ ) = ∫ c c + δ ∫ a a + δ f X , Y ( x , y ) d x d y ≈ f X , Y ( a , c ) ⋅ δ 2 P(a\leq X\leq a+\delta,c\leq Y\leq c+\delta)=\int_c^{c+\delta}\int_a^{a+\delta}f_{X,Y}(x,y)dxdy\approx f_{X,Y}(a,c)\cdot\delta^2 P(a≤X≤a+δ,c≤Y≤c+δ)=∫cc+δ∫aa+δfX,Y(x,y)dxdy≈fX,Y(a,c)⋅δ2so we can view f X , Y ( a , c ) f_{X,Y}( a, c) fX,Y(a,c) as the “probability per unit area” in the vicinity of ( a , c ) (a, c) (a,c). - The marginal PDF f X f_X fX of X X X is given by
f X ( x ) = ∫ − ∞ ∞ f X , Y ( x , y ) d y f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy fX(x)=∫−∞∞fX,Y(x,y)dy
Two-Dimensional Uniform PDF
- Let us fix some subset S S S of the two-dimensional plane. The corresponding uniform joint PDF on S S S is defined to be
- For any set A ⊂ S A \subset S A⊂S. the probability that ( X , Y ) (X, Y) (X,Y) lies in A A A is
P ( ( X , Y ) ∈ A ) = ∫ ∫ ( x , y ) ∈ A f X , Y ( x , y ) d x d y = a r e a o f A a r e a o f S P((X,Y)\in A)=\int\int_{(x,y)\in A}f_{X,Y}(x,y)dxdy=\frac{area\ of\ A}{area\ of\ S} P((X,Y)∈A)=∫∫(x,y)∈AfX,Y(x,y)dxdy=area of Sarea of A
Example 3.11. Buffon’s Needle.
A surface is ruled with parallel lines, which are at distance d d d from each other. Suppose that we throw a needle of length l l l on the surface at random. What is the probability that the needle will intersect one of the lines?
SOLUTION
- We assume here that l < d l < d l<d so that the needle cannot intersect two lines simultaneously. Let X X X be the vertical distance from the midpoint of the needle to the nearest of the parallel lines. and let θ \theta θ be the acute angle formed by the axis of the needle and the parallel lines. We model the pair of random variables ( X , Θ ) (X,\Theta) (X,Θ) with a uniform joint PDF over the rectangular set { ( x , θ ) ∣ 0 ≤ x ≤ d / 2 , 0 ≤ θ ≤ π / 2 } \{(x,\theta)|0\leq x\leq d/2,0\leq\theta\leq\pi/2\} {
(x,θ)∣0≤x≤d/2,0≤θ≤π/2}, so that
The needle will intersect one of the lines if and only if
X ≤ l 2 s i n Θ X\leq\frac{l}{2}sin\Theta X≤2lsinΘso the probability of intersection is
P ( X ≤ l 2 s i n Θ ) = ∫ ∫ x ≤ l 2 s i n θ f X , Θ ( x , θ ) d x d θ = 4 π d ∫ 0 π / 2 ∫ 0 ( l / 2 ) s i n θ d x d θ = 4 π d ∫ 0 π / 2 ( l / 2 ) s i n θ d θ = 2 l π d \begin{aligned}P(X\leq\frac{l}{2}sin\Theta)&=\int\int_{x\leq\frac{l}{2}sin\theta}f_{X,\Theta}(x,\theta)dxd\theta \\&=\frac{4}{\pi d}\int_0^{\pi/2}\int_{0}^{(l/2)sin\theta}dxd\theta \\&=\frac{4}{\pi d}\int_0^{\pi/2}(l/2)sin\theta d\theta \\&=\frac{2l}{\pi d} \end{aligned} P(X≤2lsinΘ)=∫∫x≤2lsinθfX,Θ(x,θ)dxdθ=πd4∫0π/2∫0(l/2)sinθdxdθ=πd4∫0π/2(l/2)sinθdθ=πd2l
Problem 16.
Consider the following variant of Buffon’s needle problem, which was investigated by Laplace. A needle of length l l l is dropped on a plane surface that is partitioned in rectangles by horizontal lines that are a a a apart and vertical lines that are b b b apart. Suppose that the needle’s length l l l satisfies l < a l < a l<a and l < b l < b l<b. What is the expected number of rectangle sides crossed by the needle? What is the probability that the needle will cross at least one side of some rectangle?
SOLUTION
- Let A A A be the event that the needle will cross a horizontal line, and let B B B be the probability that it will cross a vertical line. From the analysis of Example 3.11, we have that
P ( A ) = 2 l π a , P ( B ) = 2 l π b P(A) =\frac{2l}{\pi a},\ \ \ \ P(B) =\frac{2l}{\pi b} P(A)=πa2l, P(B)=πb2lThe expected number of crossed lines is
P ( A ) + P ( B ) = 2 l ( a + b ) π a b P(A) + P(B) =\frac{2l(a+b)}{\pi ab} P(A)+P(B)=πab2l(a+b) - The probability that at least one line will be crossed is
P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) P(A \cup B) = P(A) + P(B)-P(A \cap B) P(A∪B)=P(A)+P(B)−P(A∩B)Let X X X (or Y Y Y) be the distance from the needle’s center to the nearest horizontal (or vertical) line. Let Θ \Theta Θ be the angle formed by the needle’s axis and the horizontal lines. We have
P ( A ∩ B ) = P ( X ≤ l s i n Θ 2 , Y ≤ l c o s Θ 2 ) P(A \cap B) = P(X\leq\frac{lsin\Theta}{2},Y\leq\frac{lcos\Theta}{2}) P(A∩B)=P(X≤2lsinΘ,Y≤2lcosΘ)We model the triple ( X , Y , Θ ) (X, Y, \Theta) (X,Y,Θ) as uniformly distributed over the set of all ( x , y , θ ) (x, y,\theta) (x,y,θ) that satisfy 0 ≤ x ≤ a / 2 , 0 ≤ y ≤ b / 2 0 \leq x \leq a/2, 0\leq y\leq b/2 0≤x≤a/2,0≤y≤b/2, and 0 ≤ θ ≤ π / 2 0\leq \theta\leq \pi/2 0≤θ≤π/2. Hence, within this set, we have
f X , Y , Θ ( x , y , θ ) = 8 π a b f_{X,Y,\Theta}(x, y,\theta) =\frac{8}{\pi ab} fX,Y,Θ(x,y,θ)=πab8The probability P ( A ∩ B ) P(A\cap B) P(A∩B) is
Thus we have
P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) = 2 l π a + 2 l π b − l 2 π a b = l π a b ( 2 ( a + b ) − l ) P(A \cup B) = P(A) + P(B)-P(A \cap B)=\frac{2l}{\pi a}+\frac{2l}{\pi b}-\frac{l^2}{\pi ab}=\frac{l}{\pi ab}(2(a+b)-l) P(A∪B)=P(A)+P(B)−P(A∩B)=πa2l+πb2l−πabl2=πabl(2(a+b)−l)
Problem 17. Estimating an expected value by simulation using samples of another random variable.
Let Y 1 , . . . , Y n Y_1 ,...,Y_n Y1,...,Yn be independent random variables drawn from a common and known PDF f Y f_Y fY. Let S S S be the set of all possible values of Y i Y_i Yi, S = { y ∣ f Y ( y ) > 0 } S = \{y| f_Y(y) >0\} S={
y∣fY(y)>0}. Let X X X be a random variable with known PDF f X f_X fX, such that f X ( y ) = 0 f_X(y) = 0 fX(y)=0, for all y ∉ S y\notin S y∈/S. Consider the random variable
Z = 1 n ∑ i = 1 n Y i f X ( Y i ) f Y ( Y i ) Z=\frac{1}{n}\sum_{i=1}^nY_i\frac{f_X(Y_i)}{f_Y(Y_i)} Z=n1i=1∑nYifY(Yi)fX(Yi)Show that
E [ Z ] = E [ X ] E[Z] = E[X] E[Z]=E[X]
SOLUTION
We have
E [ Y i f X ( Y i ) f Y ( Y i ) ] = ∫ S y f X ( y ) f Y ( y ) f Y ( y ) d y = ∫ S y f X ( y ) d y = E [ X ] E[Y_i\frac{f_X(Y_i)}{f_Y(Y_i)}]=\int_S y\frac{f_X(y)}{f_Y(y)}f_Y(y)dy=\int_Syf_X(y)dy=E[X] E[YifY(Yi)fX(Yi)]=∫SyfY(y)fX(y)fY(y)dy=∫SyfX(y)dy=E[X]Thus,
E [ Z ] = 1 n ∑ i = 1 n E [ Y i f X ( Y i ) f Y ( Y i ) ] = 1 n ∑ i = 1 n E [ X ] = E [ X ] E[Z]=\frac{1}{n}\sum_{i=1}^nE[Y_i\frac{f_X(Y_i)}{f_Y(Y_i)}]=\frac{1}{n}\sum_{i=1}^nE[X]=E[X] E[Z]=n1i=1∑nE[YifY(Yi)fX(Yi)]=n1i=1∑nE[X]=E[X]
Joint CDFs
联合分布函数
- If X X X and Y Y Y are two random variables associated with the same experiment, we define their joint CDF by
F X , Y ( x , y ) = P ( X ≤ x , Y ≤ y ) F_{X,Y}(x,y)=P(X\leq x,Y\leq y) FX,Y(x,y)=P(X≤x,Y≤y) - In particular, if X X X and Y Y Y are described by a joint PDF f X , Y f_{X,Y} fX,Y, then
F X , Y ( x , y ) = P ( X ≤ x , Y ≤ y ) = ∫ − ∞ x ∫ − ∞ y f X , Y ( s , t ) d t d s F_{X,Y}(x,y)=P(X\leq x,Y\leq y)=\int_{-\infty}^x\int_{-\infty}^yf_{X,Y}(s,t)dtds FX,Y(x,y)=P(X≤x,Y≤y)=∫−∞x∫−∞yfX,Y(s,t)dtdsConversely, the PDF can be recovered from the CDF by differentiating:
f X , Y ( x , y ) = ∂ 2 F X , Y ( x , y ) ∂ x ∂ y f_{X,Y}(x,y)=\frac{\partial^2F_{X,Y}(x,y)}{\partial x\partial y} fX,Y(x,y)=∂x∂y∂2FX,Y(x,y)
Expectation
E [ g ( X , Y ) ] = ∫ − ∞ ∞ ∫ − ∞ ∞ g ( x , y ) f X , Y ( x , y ) d x d y E[g(X,Y)]=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)dxdy E[g(X,Y)]=∫−∞∞∫−∞∞g(x,y)fX,Y(x,y)dxdy
- As an important special case, for any scalars a , b a, b a,b, and c c c, we have
E [ a X + b Y + x ] = a E [ X ] + b E [ Y ] + c E[aX+bY+x]=aE[X]+bE[Y]+c E[aX+bY+x]=aE[X]+bE[Y]+c