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Exercise 1. Let X be a random variable such that Find a, b, and c, and the PMF of X. Justify your answer

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Answer #1

Let,x be a random variable such that

Mx(s)=a+be2s+ce4s.

E(X)=3

Var(X)=2

To find a,b, and c and the PMF of X.

We know that Mx(s) is actually the moment generating function of X and is defined as

Mx(s)

=E(esX)

=sum(esx.P(X=x))........(1)

Now,

MX(s)=ae0s+be2s+ce4s.........(2)

Comparing (1) and (2),we can say that

X takes value 0 with probability a.

X takes value 2 with probability b.

X takes value 4 with probability c.

So,Equation 1 is a+b+c=1,as X takes only these 3 values 0,2,4.

Now,E(X)=3

or,(0.a)+(2.b)+(4.c)=3

or,2b+4c=3

This is Equation 2.

Now,

Var(X)=E(X-3)2

=a*(-3)2+b*(-1)2+c*(1)2

=9a+b+c

=2

This is equation 3.

Solving these 3 equations,we get

a=1/8

b=1/4

c=5/8

Thus , PMF of X is

P(X=x)

=1/8 when x=0

=1/4    when x=2

=5/8 when x=4

Answer:-

a=1/8

b=1/4

c=5/8

And, PMF of X is

P(X=x)

=1/8 when x=0

=1/4    when x=2

=5/8 when x=4

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