Suppose that PX 2 070PX 19 020 PX 30 010 and also that

Suppose that P(X = 2) = 0.70,P(X = 19) = 0.20, P(X = 30) = 0.10, and also that Y = |X - 14|. (a) Give the PMF of Y. (b) Using (a), give EY. (c) Using (a) and (b), give varY.

A discrete random variable X has a symmetric distribution if its nonzero possible values occur in pairs: x and -x. That is, if x <>(not equal to) 0 is a possible value, then so is -x. Further, it is also the case that p(-x) = p(x). Show the following: if X is a symmetric random variable, then for any positive value b,

P(X >=b) = P(X =< -b).

Suppose that the possible values of X are x1, x2, x3 and
P(X =x1) = 3

Solution

P(X = 2) = 0.70,P(X = 19) = 0.20, P(X = 30) = 0.10

E(x) = 1.4+3.8+3 = 8.2

E(X^2) = 2.8+72.2+90= 165

Var (X) = 165-67.24 = 97.76

Var (|x-14|) = Var (x) =97.76

Hence var (y) = 97.76

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To show that P(X >=b) = P(X =< -b). for a symmetric variable

Let x=0 be the axis of symmetry.

Then prob between b and 0 and -b and 0 are equal

i.e. P(-b<x<0) = P(0<x<b)

Also x=0 divides the total probability on either side to 1/2

Hence P(x<-b) = 0.5-P(-b<x<0) =0.5-P(0<x<b)=P(X>b)

Hence proved.

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Let P(x1) = k. Then P(X2) = k/3 and P(X3) = k/4

Total prob =1 gives k+k/3+k/4 =1

k = 12/19

Or P(x1) = 12/19 P(x2) = 4/19 and P(X3) = 3/19

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mSe = E(a-x)^2 = a^2-2aE(x) +E(x^2)

Find first derivative

2a-2E(X) =0 gives a = E(x)

So the best estimate for a is mean of x.