Random variable $X$ is exponentially distributed with the parameter $\lambda$ equal to $0.5$. Define also $Y = 1 - 2X$
Find $E(Y)$ , Var(Y) and the moment generating function of Y.
I have
$f_x(X)= 0.5*e^{-0.5x}$
I think that I have to use $E(Y) = \int_{x=0}^{\infty}y*f_y(y)$ but I don't know how to find $f_y(y)$
How do I use $Y = 1 - 2X$ to find $f_y(y)$?
Since $X$ is exponentially distributed, its expected value is given by $\mathbb{E}[X] = 1/\lambda = 2$, see wikipedia on the right below the two figures. The expected value operator is linear, see wikipedia. So, we have
\begin{equation} \mathbb{E}[Y] = \mathbb{E}[1 - 2X] = \mathbb{E}[1] + \mathbb{E}[-2X] = \mathbb{E}[1] - 2\mathbb{E}[X] = 1 - 2\mathbb{E}[X] = 1 - 2 \cdot 2 = -3. \end{equation}
Can you do $\operatorname{Var}(Y) = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2$ yourself?
We define the moment-generating function of $Y$ as $M_Y(t)$. It is given by
\begin{equation} M_Y(t) = \mathbb{E}[\mathrm{e}^{tY}] = \mathbb{E}[\mathrm{e}^{t(1 - 2X)}] = \mathbb{E}[\mathrm{e}^{t} \mathrm{e}^{-2tX}] = \mathbb{E}[\mathrm{e}^{t}] \mathbb{E}[\mathrm{e}^{-2tX}]. \end{equation}
If I give you the hint that $\mathbb{E}[g(Y)] = \int_{0}^\infty g(y) f_Y(y) \,\mathrm{d}y$, where $f_Y(y)$ is the probability density function of $Y$, can you also solve for the moment generating function of $Y$?
Update
We have $\mathbb{E}[X^2] = 2/\lambda^2 = 2/(0.5)^2 = 8$. Thus,
\begin{align} \mathbb{E}[Y^2] = \mathbb{E}[(1 - 2X)^2] = \mathbb{E}[1 - 4X + 4X^2] &= \mathbb{E}[1] - 4 \mathbb{E}[X] + 4 \mathbb{E}[X^2] \\ &= 1 - 4\cdot 2 + 4 \cdot 8 = 25. \end{align}
So,
\begin{equation} \operatorname{Var}(Y) = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2 = 25 - (-3)^2 = 16. \end{equation}
Continuing for the moment-generating function:
\begin{equation} M_Y(t) = \mathbb{E}[\mathrm{e}^{t}] \mathbb{E}[\mathrm{e}^{-2tX}] = \mathrm{e}^t \mathbb{E}[\mathrm{e}^{-2tX}] = \mathrm{e}^t \int_{x = 0}^\infty \mathrm{e}^{-2tx} f_X(x) \, \mathrm{d}x, \end{equation}
where $f_X(x)$ is the probability density function of $X$ and thus satisfies $f_X(x) = \lambda \mathrm{e}^{-\lambda x}$. Substituting yields
\begin{equation} M_Y(t) = \mathrm{e}^t \int_{x = 0}^\infty \mathrm{e}^{-2tx} \lambda \mathrm{e}^{-\lambda x} \, \mathrm{d}x = \lambda \mathrm{e}^t \int_{x = 0}^\infty \mathrm{e}^{-x(2t+\lambda)} \, \mathrm{d}x = \frac{\lambda \mathrm{e}^t }{2t + \lambda}. \end{equation}
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.
$F_Y(y)=P(Y\leq y)=P(1-2X\leq y)=P(X\geq0.5-0.5y)=1-F_X(0.5-0.5y)$
Note that the last equality demands that $F_X$ is continuous.
Differentating on both sides gives $f_Y$ on LHS and an expression in $f_X$ on RHS.
