Answer to Question #137852 in Statistics and Probability for AJAY

Question

1.Let X be a random variable such that
P(X= -2)= P(X= -1), P(X= 2)= P(X= 1) and
P(X> 0)=P(X< 0)= P(X= 0).
Obtain the probability mass function of X and its distribution function.
2.A.random variable X assumes the values -3, -2, -1,0, 1,~, 3 such that
P(X= -3)= P(x= -2)= P(X= -1),
P(X= 1)= P(X= 2)= P(X= 3),
and P ( X= 0) = P ( X > 0) = P ( X < 0),
Obtain the probability mass function of X and 'its distribution function, and find
further the probability mass function of Y = 2X^2 + 3X + 4.

Answer 1

1. We supppose that $X$ takes values $-m,-m+1,\ldots,-1,0,1,\ldots,n-1,n$ with probabilities $p_{-m},p_{-m+1},..,p_{-1},p_0,p_1,p_2,\ldots,p_{n-1},p_n$ . It has to be satisfied: $\sum_{i=-m}^{n}p_i=1$ . From the conditions we receive that $\sum_{i=-m}^{-1}p_i=p_0=\sum_{i=1}^{n}p_i$ and $p_{-2}=p_{-1}, p_1=p_2$. The distribution function is then given by $F_X(x)=P(X\leq x)=\sum_{i=-m}^xp_i$.
2. We denote $p_1=P(X=-3)=P(X=-2)=P(X=-1),$ $p_2=P(X=1)=P(X=2)=P(X=3),$ $p_3=P(X=0)=3p_1=3p_2$ . From the latter we find that $p_1=p_2$ . From Equality $3p_1+3p_2+p_3=6p_1+3p_1=1$ yields $p_1=\frac19$ . Thus,

$P(X=-3)=P(X=-2)=P(X=-1)=\frac19,$

$P(X=1)=P(X=2)=P(X=3)=\frac19,$ $p_3=\frac13$

The distribution function is then given by $F_X(x)=P(X\leq x)=\sum_{i=-m}^xp_i$ . The function $Y$ takes values 13, 6, 0, 4, 9, 18, 31. The probability mass function can be taken from the probabilities:

$P(X=13)=P(X=6)=P(X=0)=\frac19,$ $P(X=4)=\frac13$ ,

$P(X=9)=P(X=18)=P(X=31)=\frac19.$