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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 $f_y(y)$?

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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}

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  • $\begingroup$ Thanks! Var[X] = 4 so Var[Y] = Var[1] - 2Var[X] = 1-2*4 = -7 is this correct? I will also try the last one. $\endgroup$ – Marvin Jun 29 '15 at 11:50
  • $\begingroup$ @Marvin that is not correct. In general $\text{Var}(a+bX)=b^2\text{Var}(X)$ if $a$ and $b$ are constants. A variance cannot be negative. $\endgroup$ – drhab Jun 29 '15 at 11:55
  • $\begingroup$ @Marvin Think about what $\mathbb{E}[Y^2]$ actually is, namely $\mathbb{E}[Y^2] = \mathbb{E}[(1 - 2X)^2] = \mathbb{E}[1 - 4X + 4X^2] = \cdots$ and use the same linearity property we just used for the expected value of $Y$. $\endgroup$ – Ritz Jun 29 '15 at 12:05
  • $\begingroup$ Since it is exponentially distributed $Var[X] = 1/(lambda^2) = 4$ and using $Var(a+bX)=b^2*Var(X)$ we have $4*Var(X) = 16$. But how can I find $E[X^2]$ ? And can you please show me how to find the moment generating function also? I need to see it once to understand it. $\endgroup$ – Marvin Jun 29 '15 at 14:08
  • $\begingroup$ $E[X^n] = n!/lambda^n $ $\endgroup$ – Marvin Jun 29 '15 at 14:17
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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.

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