Let X and Y be two random variables with joint probability mass function:P(X=i,Y=j) = (i+j)/36, for i,j=1,2,3,4What is the marginal probability mass function of X?Review LaterP(X=i) = 5(i+1)/36, for i=1,2,3,4P(X=i) = (5+i)/36, for i=1,2,3,4P(X=i) = 5/12, for i=1,2,3,4P(X=i) = (5i+1)/36, for i=1,2,3,4
Question
Let X and Y be two random variables with joint probability mass function:P(X=i,Y=j) = (i+j)/36, for i,j=1,2,3,4What is the marginal probability mass function of X?Review LaterP(X=i) = 5(i+1)/36, for i=1,2,3,4P(X=i) = (5+i)/36, for i=1,2,3,4P(X=i) = 5/12, for i=1,2,3,4P(X=i) = (5i+1)/36, for i=1,2,3,4
Solution
The marginal probability mass function of X can be found by summing the joint probabilities for each value of X.
For X = 1, we have P(X=1) = (1+1)/36 + (1+2)/36 + (1+3)/36 + (1+4)/36 = 5/36 + 6/36 + 7/36 + 8/36 = 26/36 = 13/18.
Similarly, for X = 2, we have P(X=2) = (2+1)/36 + (2+2)/36 + (2+3)/36 + (2+4)/36 = 7/36 + 8/36 + 9/36 + 10/36 = 34/36 = 17/18.
For X = 3, we have P(X=3) = (3+1)/36 + (3+2)/36 + (3+3)/36 + (3+4)/36 = 10/36 + 11/36 + 12/36 + 13/36 = 46/36 = 23/18.
Finally, for X = 4, we have P(X=4) = (4+1)/36 + (4+2)/36 + (4+3)/36 + (4+4)/36 = 13/36 + 14/36 + 15/36 + 16/36 = 58/36 = 29/18.
Therefore, the marginal probability mass function of X is: P(X=1) = 13/18, P(X=2) = 17/18, P(X=3) = 23/18, P(X=4) = 29/18.
Similar Questions
Problem 1 (15 points). Suppose (X, Y ) are discrete random variables with joint probability massfunction given by the following table:x\y -2 0 2-2 3/16 1/16 3/160 1/16 0 1/162 3/16 1/16 3/16a. (6 points) Find the marginal pmf’s for X and Y and check if the two random variables areindependent.b. (5 points) Show that the covariance between X and Y is zero.c. (4 points) Is there any contradiction between the results in parts (a) and (b)?
Suppose that X and Y have the joint probability density function: f(x,y) = 30x^2y^2(1 - x^2 - y^2) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. What is the marginal probability density function of X?Review LaterfX(x) = 15x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 15x^2(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x^2(1 - x^2) for 0 ≤ x ≤ 1
4Q. The joint probability mass function of discrete random variables 𝑋 X and 𝑌 Y is given by: 𝑃 [ 𝑋 = 𝑥 , 𝑌 = 𝑦 ] = { 𝑘 ( 4 𝑥 + 𝑦 ) , 𝑥 = 1 , 2 , 3 and 𝑦 = 1 , 2 , 3 0 , otherwise P[X=x,Y=y]={ k(4x+y), 0, x=1,2,3 and y=1,2,3 otherwise Find the value of 𝑘 k. Then, calculate 𝑃 ( 𝑋 = 2 and 𝑌 = 3 ) P(X=2 and Y=3).
The random variables X and Y have the joint PMF: px,y(x, y)=c*(x+y)^(2) if x belongs to {1,2,4} and y belongs to {1,3} and otherwise px,y(x,y) =0. Find the value of c. Find P(Y<X) and P(Y=X).
Random variable X may take value -1 or 1; Y may take value 1, 2 or 3. The following Table shows the joint probability of random variables X and Y.Y1 2 3X -1 0.1 0.2 0.31 0.2 0.1 0.1Table 2: Joint probability distribution of X and Y. For example, the probability of X=-1 and Y=1 is 0.1. Provide solution to all questions below.(a) Marginal probability distribution of X(b) The expected value, variance and standard deviation of X. (c) Marginal probability distribution of Y(d) The expected value, variance and standard deviation of Y. (e) Conditional probability distribution of Y given X=-1.(f) The conditional expected value, conditional variance and conditional standard deviation of Y
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.