Answer/ Explanation:
Since X is exponentially distributed, its expected value is given by E[X]=1/λ=2.
Therefore, E[Y]=E[1−2X]=E[1]+E[−2X]=E[1]−2E[X]=1−2E[X]=1−2⋅2=−3.
Hence,
We define the moment-generating function of Y as MY(t). It is given by
MY(t)=E[etY]=E[et(1−2X)]=E[ete−2tX]=E[et]E[e−2tX].
If I give you the hint that E[g(Y)]=∫∞0g(y)fY(y)dy, where fY(y) is the probability density function of Y, can you also solve for the moment generating function of Y?
We have E[X2]=2/λ2=2/(0.5)2=8. Thus,
E[Y2]=E[(1−2X)2]=E[1−4X+4X2]=E[1]−4E[X]+4E[X2]=1−4⋅2+4⋅8=25.
So,
Var(Y)=E[Y2]−E[Y]2=25−(−3)2=16.
Continuing for the moment-generating function:
MY(t)=E[et]E[e−2tX]=etE[e−2tX]=et∫∞x=0e−2txfX(x)dx,
where fX(x) is the probability density function of X and thus satisfies fX(x)=λe−λx. Substituting yields
MY(t)=et∫∞x=0e−2txλe−λxdx=λet∫∞x=0e−x(2t+λ)dx=λet2t+λ.
It is also good to note that
If you are after expectation, variance or moment generating function of Y then it is not needed to find the PDF of Y (see the answer of Ritz).
This is not an answer on the question in the title, but one on the question in the body.
FY(y)=P(Y≤y)=P(1−2X≤y)=P(X≥0.5−0.5y)=1−FX(0.5−0.5y)
Note that the last equality demands that FX is continuous.
Differentating on both sides gives fY on LHS and an expression in fX on RHS.
