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​

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 joint pdf of two continuous random variables 𝑋X and 𝑌Y is given by𝑓𝑋𝑌(𝑥,𝑦)={4𝑥𝑦1440≤𝑥≤14,0≤𝑦≤140otherwisef XY​ (x,y)={ 14 4 4xy​ 0​ 0≤x≤14,0≤y≤14otherwise​ Are 𝑋X and 𝑌Y independent?YesNo

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]

eet the joint p.d.f. of X1 and X2 be:ℎ(𝑥1, 𝑥2) = {8𝑥1𝑥2 for 0 < 𝑥1 < 𝑥2 < 10 otherwisea) Find the joint p.d.f. of 𝑌1 = 𝑋1𝑋2and 𝑌2 = 𝑋2b) Are 𝑌1 and 𝑌2 independent? Why? (10 points

(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)