Discrete Random Variable maths help question?

The discrete random variable X has probability distribution given by





P(X=x)=k(1+x) for x=1,2,3,4,5


P(X=x)=0 otherwise





(a) Show that k=1/20


(b) Find the mean and variance of X.


(c) Given that X₁, X₂ are two independant observations of X, evaluate


P(X₁+X₂=4)


(d) The random variable Y is defined by


Y=2X+3


Find the mean and varianve of Y

Discrete Random Variable maths help question?
a. Consider the probability mass function for X. We know that since X represents a probability, the sum of P(X=x) for all values of x must be 1. Given that P(X = x) = k(1+x) for x = 1,2,3,4,5, simply take the following sum and set it to 1:





P(X=1) + P(X=2) + P(X=3) + P(X = 4) + P(X=5) = 1


2k + 3k + 4k +5k + 6k = 1


20k = 1


k = 1/20








b. The mean of X is given by its expected value E[X] which is the sum of xP(X=x) for all values of x.





1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)+5P(X=5...


(1/20)(2+2*3+3*4+4*5+5*6)


(1/20)(2+6+12+20+30)


7/2





The variance of X is the sum of x^2P(X=x) for all values of x. The calculation follows similarly as above and gives a variance of 14.





c. There are only a couple pairs of numbers that will equal 4 which have positive probability. These pairs are X1 = 1, X2 = 3; X1 = 2, X2 = 2; X1 = 3; X2 = 1. Just find the probability for each pair and sum them up. Since the observations are independent P(X1 = x, X2 = y) = P(X1=x)P(X2=y) which will make your calculations fairly straightforward.





d. If Y = 2X+3, then by linearity of expectation E[Y] = E[2X+3] = 2E[X] + 3 = 10