Find pdf of $X = 5Z_1 - Z_2$, where $Z_1$, $Z_2$ independent $$X = 5Z_1 - Z_2,$$ where $Z_1$ and $Z_2$ are independent.
Find pdf of $X$.
My approach is to find the cdf $\to$ differentiate $\to$ pdf
cdf:
$$\begin{align}F_X(x) 
&= P(5Z_1 - Z_2 \le x)\\
&= P(Z_1 \le (X+Z_2)/5)\\
&= E[F_{Z_1}(X+Z_2)/5]\end{align}$$
Let $f_X(x) = F'X(x)$ where $f_X(x)$ is the pdf of $X$
$f_X(x)=E[f_{Z_1} (X+Z_2)/5 * 1/5]$
 A: Your idea is good, but the calculations are all wrong.
$$
F_X(x) = P\{X \leq x\} = P\{5Z_1-Z_2 \leq x\}
= \int_{z_1 = -\infty}^\infty \int_{z_2 = 5z_1-x}^\infty f_{1,2}(z_1,z_2)
\,\mathrm dz_2 \,\mathrm dz_1\\
$$
If you differentiate $F_X(x)$ with respect to $x$ to get the density
function $f_X(x)$, you will find that
$$\begin{align}f_X(x) &= \frac{\partial F_X(x)}{\partial x}\\
&=\frac{\partial}{\partial x} \int_{z_1 = -\infty}^\infty\left[ \int_{z_2 = 5z_1-x}^\infty f_{1,2}(z_1,z_2)
\,\mathrm dz_2\right] \,\mathrm dz_1\\
&= \int_{z_1 = -\infty}^\infty\left[ \frac{\partial}{\partial x}
\int_{z_2 = 5z_1-x}^\infty f_{1,2}(z_1,z_2)
\,\mathrm dz_2\right] \,\mathrm dz_1\\
&= \int_{-\infty}^\infty f_{1,2}(z_1,5z_1-x)\,\mathrm dz_1.
\end{align}$$
At this point, if you want, you can use the independence of
$Z_1$ and $Z_2$ to express $f_{1,2}(z_1, 5z_1-x)$ as
$f_1(z_1)f_2(5z_1-x)$ and proceed to calculate the
density of $X$ in more detail.

For the special case when $Z_1$ and $Z_2$ are independent normal random
variables, then you can use the facts that


*

*$X = 5Z_1-Z_2$ also is a normal random variable

*$E[X] =5E[Z_1] - E[Z_2]$

*$\operatorname{var}(X) = 5^2\operatorname{var}(Z_1) +
\operatorname{var}(Z_2)$
to write down the density of $X$ without doing any integration.
A: Hint: assuming that $Z_1,Z_2$ are standard normal distributed, then:


*

*You know the parameters (mean, variance) of $Z_1,Z_2$.

*Use what you know about how expectations add to work out the mean of $X$

*Use what you know about how independent variances add to work out the mean of $X$

*Use what you know about the distribution of the sum of independent normal variables to work out what the distribution of $X$ is

*Plug the parameters of $X$ (points 2,3) into the probability density for $X$ (point 4). If your probability density is a standard density, you need to look up the formula for that density in a reference book (or on wikipedia) and then substitute your values for mean $E(X)=\mu$ and variance $V(X)=\sigma^2$ into that formula.
