The joint distribution of 𝑋X and 𝑌Y is given by                             𝑓𝑋𝑌(𝑥,𝑦)=916×4𝑥+𝑦,f XY​ (x,y)= 16×4 x+y 9​ ,where 𝑇𝑋,𝑇𝑌∈{0,1,2,…}.T X​ ,T Y​ ∈{0,1,2,…}.1 pointFind the probability mass function of 𝑋+𝑌X+Y.𝑘916⋅4𝑘k 16⋅4 k 9​ (𝑘+1)916⋅4𝑘(k+1) 16⋅4 k 9​ (𝑘+1)916⋅4𝑘+1(k+1) 16⋅4 k+1 9​ 𝑘916⋅4𝑘+1k 16⋅4 k+1 9​

For the Joint PMF as shown, find each of following quantities:𝑝𝑋 𝑥 , 𝑝𝑌 𝑦 , 𝑝𝑋|𝑌 𝑥 𝑦 , 𝑝𝑌|𝑋 𝑦 𝑥 , 𝐸[𝑋|𝑌 = 3]Also find whether 𝑋 and 𝑌 are independent or not.[The graph shows 𝑝𝑋,𝑌(𝑥, 𝑦)/12]

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).

(30 points) Suppose Company 1's stock return 𝑋𝑋 is a random variable and takes three possiblevalues: {-0.1, 0.1, 0.2}. And Company 2's stock return 𝑌𝑌 is a random variable and takes twopossible values: {-0.3, 0.4}. The joint probability distribution 𝑓𝑓(𝑋𝑋, 𝑌𝑌) is given as follows:𝑓𝑓(−0.1, −0.3) = 0.1, 𝑓𝑓(0.1, −0.3) = 0.2, 𝑓𝑓(0.2, −0.3) = 0.2,𝑓𝑓(−0.1,0.4) = 0.2, 𝑓𝑓(0.1,0.4) = 0.2, 𝑓𝑓(0.2,0.4) = 0.1.Please calculate the following:(a) Marginal distributions: 𝑓𝑓𝑋𝑋(𝑥𝑥) and 𝑓𝑓𝑌𝑌(𝑦𝑦). (4 points)(b) Mean: 𝐸𝐸(𝑋𝑋) and 𝐸𝐸(𝑌𝑌). (4 points)(c) Variance: 𝑉𝑉𝑎𝑎𝑟𝑟(𝑋𝑋) and 𝑉𝑉𝑎𝑎𝑟𝑟(𝑌𝑌). (4 points)(d) Covariance: 𝐶𝐶𝐶𝐶𝐶𝐶(𝑋𝑋, 𝑌𝑌). (2 points)(e) Conditional expectations: 𝐸𝐸(𝑋𝑋|𝑌𝑌 = −0.3) and 𝐸𝐸(𝑋𝑋|𝑌𝑌 = 0.4). (8 points)(f) Conditional variances: 𝑉𝑉𝑎𝑎𝑟𝑟(𝑋𝑋|𝑌𝑌 = −0.3) and 𝑉𝑉𝑎𝑎𝑟𝑟(𝑋𝑋|𝑌𝑌 = 0.4). (8 points)

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