So the question is:
If X is a random variable with a cdf $FX (t)$, and Y is the random variable given by Y = aX + b. Express the cdf $FY (t)$ of Y in terms of $FX (t)$.
(Consider separately the cases a > 0, a < 0 and a = 0. Notice that it is not assumed that X is either a discrete or a continuous random variable! You may use limits if necessary.)
What I have so far is :
$F_X(t)=P(X\leq t)$
Now Y=aX+b
So CDF of the variable Y is $F_Y(t)=P(Y\leq t)$
i) for a>0,
$F_Y(t)=P(Y\leq t) =P(aX+b\leq t) =P(X\leq \frac{t-b}{a}) =F_X(\frac{t-b}{a})$
ii) for a<0
$F_Y(t)=P(Y\leq t) =P(aX+b\leq t) =P(aX\leq t-b)=P(X\geq \frac{t-b}{a}) [ since a<0 ]$ =$1-P(X<\frac{t-b}{a} )$
If X is a discrete random variable,
$F_Y(t)=1-P(X<\frac{t-b}{a} )=1-P(X\leq \frac{t-b}{a}-1)=1-F_X(\frac{t-b-a}{a})$
If X is a continuous random variable,
$F_Y(t)=1-P(X<\frac{t-b}{a} )=1-F_X(\frac{t-b}{a})$
iii) for a=0
$Y=aX+b=(0\times X)+b=b$
Here the variable Y has only one mass point i.e. Y=b.
$\therefore P(Y=b)=1$
$\therefore P(Y\leq b)=1\Rightarrow F_Y(b)=1$
So the CDF of the variable Y is
for a<0
$F_Y(t)=1-F_X(\frac{t-b-a}{a})$ , X is a discrete random variable.
$F_Y(t)=1-F_X(\frac{t-b}{a}) $ , X is a continuous random variable.
(for any value of t .)
for a>0
$F_Y(t)=F_X(\frac{t-b}{a})$
(for any value of t .)
for a=0
$F_Y(t)=1 $ (where t=b).
Does the solution look right? I worked out the discrete random variable and the continuous random variable parts in ii with my fellow, but I now again got lost in this part... I don't see why if x is a discrete random variable $1-P(X<\frac{t-b}{a} )=1-P(X\leq \frac{t-b}{a}-1)$ but if x is the continuous random variable it is just $1-P(X<\frac{t-b}{a} )$. Would any one explain a little maybe?
