Tutorial 4:Discrete Random Variables 2 Baoxiang Wang bxwang@cse Spring 2017 1
Tutorial 4: Discrete Random Variables 2 Baoxiang Wang bxwang@cse Spring 2017 1
Geometric Random variables The PMF of a geometric random variable is px(k)=(1-p)k-1p ·E[XX>1]=1+E[X] 2
Geometric Random Variables • The PMF of a geometric random variable is 𝑝𝑋 𝑘 = (1 − 𝑝) 𝑘−1𝑝 • 𝐸 𝑋 𝑋 > 1 = 1 + 𝐸[𝑋] 2
Geometric Random variables For k 1, P(X=klX >1)= P(X=k 2=1-p)k-1 2=_(1-p)k-1卫 (X>1) 2R2-2(1-p)k-1p 1 p(1-p)1-1-p =(1-p)k-2p=px(k-1) ·EXX>1]=∑=2k(X=kX>1) -夏p=+1p肉公n.+的 k=2 k三1 =1+E[X] 3
Geometric Random Variables • For 𝑘 > 1, 𝑃 𝑋 = 𝑘 𝑋 > 1 = 𝑃(𝑋=𝑘) 𝑃(𝑋>1) = (1−𝑝) 𝑘−1𝑝 (1−𝑝) ∞ 𝑘−1𝑝 𝑘=2 = (1−𝑝) 𝑘−1𝑝 𝑝(1−𝑝)∙ 1 1−(1−𝑝) = (1 − 𝑝) 𝑘−2𝑝 = 𝑝𝑋(𝑘 − 1) • 𝐸 𝑋 𝑋 > 1 = 𝑘𝑃 𝑋 = 𝑘 𝑋 > 1 ∞ 𝑘=2 = 𝑘𝑝𝑋(𝑘 − 1) ∞ 𝑘=2 = (𝑘 + 1)𝑝𝑋(𝑘) ∞ 𝑘=1 = 𝑘𝑝𝑋(𝑘) ∞ 𝑘=1 + 𝑝𝑋(𝑘) ∞ 𝑘=1 = 1 + 𝐸[𝑋] 3
Geometric Random variables ·In general, P=k) P(X=k X>a)= P(X>a) (1-p)k-1p (1-p)k-1p 28=a+1(1-p)k-1p 1 1 (1-p)a.T-1-p) =(1-p)k-a-1p =px(k-a) 4
Geometric Random Variables • In general, 𝑃 𝑋 = 𝑘 𝑋 > 𝑎 = 𝑃(𝑋 = 𝑘) 𝑃(𝑋 > 𝑎) = (1 − 𝑝) 𝑘−1𝑝 (1 − 𝑝) 𝑘−1𝑝 ∞ 𝑘=𝑎+1 = (1 − 𝑝) 𝑘−1𝑝 𝑝(1 − 𝑝) 𝑎∙ 1 1 − (1 − 𝑝) = (1 − 𝑝) 𝑘−𝑎−1𝑝 = 𝑝𝑋(𝑘 − 𝑎) 4
Geometric Random variables E[f(X)X a]=E[f(X +a)] ·E[f(X)IX>a] 00 =∑f)P(X=kX>a)=∑f)px(k-) k=a+1 k=a+1 00 =f(k+a)px(k)=EIf(X+a)] k三1
Geometric Random Variables • 𝐸 𝑓 𝑋 𝑋 > 𝑎 = 𝐸[𝑓(𝑋 + 𝑎)] • 𝐸 𝑓 𝑋 𝑋 > 𝑎 = 𝑓(𝑘)𝑃 𝑋 = 𝑘 𝑋 > 𝑎 ∞ 𝑘=𝑎+1 = 𝑓(𝑘)𝑝𝑋(𝑘 − 𝑎) ∞ 𝑘=𝑎+1 = 𝑓(𝑘 + 𝑎)𝑝𝑋(𝑘) ∞ 𝑘=1 = 𝐸[𝑓(𝑋 + 𝑎)] 5
Example 1:Functions of random variables Let X be a random variable that takes value from 0 to 9 with equal probability 1/10. .(a)Find the PMF of the random variable Y =X mod(3). X 0,3,6,9 1,4,7 25,8 Y 0 1 2 0=2/5 3/10 3/10 6
Example 1: Functions of random variables • Let 𝑋 be a random variable that takes value from 0 to 9 with equal probability 1/10. • (a) Find the PMF of the random variable 𝑌 = 𝑋 𝑚𝑜𝑑(3). 6 X 0,3,6,9 1,4,7 2,5,8 Y 0 1 2 P 4 10 = 2/5 3/10 3/10
Example 1:Functions of random variables Let X be a random variable that takes value from 0 to 9 with equal probability 1/10. .(b)Find the PMF of the random variable Y 5 mod(X+1). X 0 3 4 5 6 8 0 2 1 0 5 5 5 5 5 0 1 2 5 品-5 品-15 1/10 0=1/2 7
Example 1: Functions of random variables • Let 𝑋 be a random variable that takes value from 0 to 9 with equal probability 1/10. • (b) Find the PMF of the random variable 𝑌 = 5 𝑚𝑜𝑑(𝑋 + 1). 7 X 0 1 2 3 4 5 6 7 8 9 Y 0 1 2 1 0 5 5 5 5 5 Y 0 1 2 5 P 2 10 = 1/5 2 10 = 1/5 1/10 5 10 = 1/2
Example 2:Expectation,Mean,and Variance Let X be a random variable with PMF x2 Px(x)= ifx=-3,-2,-1,0,1,2,3 0, otherwise ·(a)Find a and E[X] a- x2=2×(1+4+9)=28 EW=∑ x 280 X=-3 8
Example 2: Expectation, Mean, and Variance • Let 𝑋 be a random variable with PMF 𝑝𝑋 𝑥 = 𝑥 2 𝑎 , 𝑖𝑓 𝑥 = −3, −2, −1,0,1,2,3 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 • (a) Find 𝑎 and 𝐸[𝑋] 𝑎 = 𝑥 2 3 𝑥=−3 = 2 × 1 + 4 + 9 = 28 𝐸 𝑋 = 𝑥 ∙ 𝑥 2 28 3 𝑥=−3 = 0 8
Example 2:Expectation,Mean,and Variance Let X be a random variable with PMF x2 Px(x)= ifx=-3,-2,-1,0,1,2,3 0, otherwise .(b)What is the PMF of the random variable Z =(X-E[X])2? Z=X2 X 0 -1,1 2,2 3,3 Z 0 1 4 9 0 1 42 9 9 2×28=1/14 2×28=7 2×28=14 9
Example 2: Expectation, Mean, and Variance • Let 𝑋 be a random variable with PMF 𝑝𝑋 𝑥 = 𝑥 2 𝑎 , 𝑖𝑓 𝑥 = −3, −2, −1,0,1,2,3 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 • (b) What is the PMF of the random variable 𝑍 = (𝑋 − 𝐸[𝑋]) 2 ? 𝑍 = 𝑋 2 9 X 0 -1,1 -2,2 -3,3 Z 0 1 4 9 P 0 2 × 1 28 = 1/14 2 × 4 28 = 2 7 2 × 9 28 = 9 14
Example 2:Expectation,Mean,and Variance Let X be a random variable with PMF x2 Px(x)= ifx=-3,-2,-1,0,1,2,3 (0, otherwise .(c)Using the result from part(b),find the variance of X. 2 999 var(X)=E[Z]=1×14+4×气+9× 14=14 Z 0 1 4 9 0 2×1/28=1 2×4/28=2 2×9/28=9 /14 /7 /14 10
Example 2: Expectation, Mean, and Variance • Let 𝑋 be a random variable with PMF 𝑝𝑋 𝑥 = 𝑥 2 𝑎 , 𝑖𝑓 𝑥 = −3, −2, −1,0,1,2,3 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 • (c) Using the result from part (b), find the variance of 𝑋. 𝑣𝑎𝑟 𝑋 = 𝐸 𝑍 = 1 × 1 14 + 4 × 2 7 + 9 × 9 14 = 99 14 10 Z 0 1 4 9 P 0 2 × 1/28 = 1 /14 2 × 4/28 = 2 /7 2 × 9/28 = 9 /14
