Let X be a random variable such that
P [X=-2) = P[X=-I]P[X =2] = P[X=1]
and P[x>0 ] = p [x< 0]= P[x=0] obtain the probability mass function of x
Answers
Answer:
We denote p_1=P(X=-3)=P(X=-2)=P(X=-1),p
1
=P(X=−3)=P(X=−2)=P(X=−1), p_2=P(X=1)=P(X=2)=P(X=3),p
2
=P(X=1)=P(X=2)=P(X=3), p_3=P(X=0)=3p_1=3p_2p
3
=P(X=0)=3p
1
=3p
2
. From the latter we find that p_1=p_2p
1
=p
2
. From Equality 3p_1+3p_2+p_3=6p_1+3p_1=13p
1
+3p
2
+p
3
=6p
1
+3p
1
=1 yields p_1=\frac19p
1
=
9
1
. Thus,
P(X=-3)=P(X=-2)=P(X=-1)=\frac19,P(X=−3)=P(X=−2)=P(X=−1)=
9
1
,
P(X=1)=P(X=2)=P(X=3)=\frac19,P(X=1)=P(X=2)=P(X=3)=
9
1
, p_3=\frac13p
3
=
3
1
The distribution function is then given by F_X(x)=P(X\leq x)=\sum_{i=-m}^xp_iF
X
(x)=P(X≤x)=∑
i=−m
x
p
i
. The function YY takes values 13, 6, 0, 4, 9, 18, 31. The probability mass function can be taken from the probabilities:
P(X=13)=P(X=6)=P(X=0)=\frac19,P(X=13)=P(X=6)=P(X=0)=
9
1
, P(X=4)=\frac13P(X=4)=
3
1
,
P(X=9)=P(X=18)=P(X=31)=\frac19.P(X=9)=P(X=18)=P(X=31)=
9
1
.