Question

Let `X` be a random variable with probability density distribution given by `f(x)={(x if 0<=x<=1),(1 if 1<x<=1.5),(0 quad otherwise):}`.

Find the probability density function of `Y=10X-4`.

Answer

Let CDF of `X` is `F(x)` and CDF of `Y` is `F(y)` then since `Y=10X-4` we have that `F(y)=P(Y<=y)=P(10X-4<=y)=P(X<=(y+4)/10)`.

Now, consider two cases.

Case 1. `0<=(y+4)/10<=1` or `-4<=y<=6`.

`P(X<=(y+4)/10)=int_0^((y+4)/10)xdx=1/2 x^2|_0^((y+4)/10)=1/2 ((y+4)/10)^2`.

Case 2. `1<(y+4)/10<=1.5` or `6<y<=11`.

`P(X<=(y+4)/10)=int_0^1 xdx+int_1^((y+4)/10)1*dx=1/2 x^2|_0^1+x|_1^((y+4)/10)=1/2+(y+4)/10-1=(y+4)/10-1/2`.

So, `F(y)={(0 if y<-4),(1/2((y+4)/10)^2 if -4<=y<=6),((y+4)/10-1/2 if 6<y<=11),(1 if y>11):}`.

To obtain PDF, we just differentiate CDF:

`f(y)={(0 if y<-4),(1/100 (y+4) if -4<=y<=6),(1/10 if 6<y<=11),(0 if y>11):}`.

Thus,

`f(y)={(1/100 (y+4) if -4<=y<=6),(1/10 if 6<y<=11),(0 quad otherwise):}`.