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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