$$X = 5Z_1 - Z_2,$$ 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 $X$

$f_X(x)=E[f_{Z_1} (X+Z_2)/5 * 1/5]$

  • $\begingroup$ I made some edits to your post to improve the formatting. Can you please check it over and make sure that it's what you meant? By the way thanks a lot for including your approach. $\endgroup$
    – TooTone
    Commented Feb 17, 2014 at 1:22
  • $\begingroup$ Thank you TooTone, for your editing. im quite new to here. after I got E(X) = 0 and Var(X) = 26 How can I continue to get the probability density function ? $\endgroup$
    – augee
    Commented Feb 17, 2014 at 10:48
  • $\begingroup$ That's correct if you assume that Z1 and Z2 are standard normal. Are you told that they are standard normal? $\endgroup$
    – TooTone
    Commented Feb 17, 2014 at 11:05

2 Answers 2


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.

  • $\begingroup$ i got this, thanks for the assumption too ! $\endgroup$
    – augee
    Commented Feb 17, 2014 at 11:02

Hint: assuming that $Z_1,Z_2$ are standard normal distributed, then:

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

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

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

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

  5. 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.


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