I am not able to understand the difference between the joint density function and density function for a random variable $Z = X_1 + X_2$, where $X_1, X_2$ are uniform random variables in $[0,1]$.

I think joint density in this case is $f_{X_1, X_2}\left(x_1,x_2\right) = 1$ (reference)

Likewise the density function $f_Z$ is defined as convolution of $f_{X_1}$ and $f_{X_2}$ (reference: page 8)

Could someone please explain the difference between the two?

  • $\begingroup$ When there is more than a single variable we use the notion of joint probability density function.Taking a look at this link may help you to get the answer of your question. probabilitycourse.com/chapter5/5_2_1_joint_pdf.php $\endgroup$
    – user53119
    Feb 13, 2016 at 22:00
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    $\begingroup$ Asked 4 hours earlier on math.SE and answered fully in a comment there by Math.SE Moderator Andre Nicolas. $\endgroup$ Feb 14, 2016 at 15:28

2 Answers 2


Following up on Glen_b's answer, and in an attempt to dumb it down a bit more, the following illustrations shows how the bivariate or joint pdf of $X$ and $Y$, both independent and standard uniform distributions $U(0,1)$, forms the basis to obtain the pdf of the variable $Z$, defined as the sum of $X + Y.$

For every value of $Z,$ the probability of getting a quantile lower than $Z,$ i.e. the distribution or cdf of $Z$ will be given by the integral from $-\infty$ to the line $z = x + y$ on the $X$ axis, and from $-\infty$ to $+\infty$ on the $Y$ axis. Luckily, the vast majority of the $XY$ plane corresponds to a joint pdf of $0$.

The idea is that by obtaining the cdf, i.e. $F_Z(z)$, of $Z$ we can arrive at the cdf, i.e. $f_Z(z)$, by differentiating:

enter image description here

In general, the pdf of the sum of two independent variables $X$ and $Y$ can be proven to correspond to the convolution of $f_X(z)$ and $f_Y(z)$ at $z$, i.e.

$$f_Z(z)=(f * g)(z)= \int_{-\infty}^{+\infty}f_X(z-y)\,f_Y(y)\,\mathrm dy.$$

Algebraically, the joint density can be expressed by two indicator variables:



$$f_Z(z)=\int_0^z f_{X}(z-y)f_{Y}(y)\,\mathrm dy=\int_0^z 1\!_{\{(z-y)\,\in\, [0,1]\}}\,1\!_{\{(y\,\in\, [0,1]\}}\,\mathrm dy$$

$Z=X+Y$ can reach $2$ inviting the following partition:

  1. If $0\leq z\leq1$,

and noting that $f_Y(y)=1$ if $0\leq y \leq 1$, and $0$ otherwise,

$$\begin{align} f_Z(z)&=\int_0^1 1\!_{\{(z-y)\in [0,1]\}}\,\mathrm dy\\[2ex] &=\int_0^z \mathrm dy \\[2ex] &= z \end{align}$$

  1. If $1\leq z\leq 2$,

and given that we are integrating over $y$, and the integrand is $1$ only when $0\leq z−y \leq 1$ (rearranged: $z−1\leq y\leq z$):

$$f_Z(z)= \int_{z-1}^{1} \mathrm dy = 2-z$$

explaining the triangular appearance of the $\text{pdf}$ of $Z$.

  • $\begingroup$ +1. This GIF conveys the geometric intuition for how to think about the transformations beautifully. Out of interest, what did you use to create it? $\endgroup$
    – microhaus
    May 16, 2021 at 20:23
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    $\begingroup$ @microhaus I did it with Geogebra. I find nothing like it in terms of esthetic options when you are not working with simulated data or data sets. $\endgroup$ May 16, 2021 at 21:12
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    $\begingroup$ Thank you for the software recommendation. When I've computed these transformations previously, this is what I'd visualise in my head or in 2D on paper. It's somewhat of a sight to see a 'thought artefact' coming to life as it were. And at least for me, it's pedgagogically much more appealing than the terse algebraic descriptions supplied in many statistics and probability reference textbooks. $\endgroup$
    – microhaus
    May 16, 2021 at 21:32
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    $\begingroup$ A minor notational issue - you should really notate the convolution as $(f * g)(z)$ rather than $f(z) * g(z)$. The convolution acts on the functions, not on the elements of the codomains. $\endgroup$ May 17, 2021 at 15:09

If you don't write down the support, you may not see what's going on -- but as soon as you do, it's a lot clearer.

I am not able to understand the difference between the joint density function and density function for a random variable Z = x1 + x2 where x1, x2 are uniform rvs in [0,1].

Note that $f_{X_1}(x_1)=1$ for $0<x_1<1$ and $0$ elsewhere; similarly for $X_2$.

The joint density is bivariate - the density is a surface.

I think joint density in this case is f(x1,x2) = 1

So, assuming independence, the joint density will be: $f(x_1,x_2) = f_{X_1}(x_1)\, f_{X_2}(x_2)= 1 \times 1=1$ on the unit square and $0$ elsewhere.

![enter image description here

(At least, "bivariate uniform under independence")

Likewise the density function of z is defined as convolution of x1 and x2

It is the convolution if they're independent, yes.

The sum of a pair of quantities is a single quantity -- the sum of a pair of random variables is a univariate random variable.

The density function of the sum of independent variables goes from the sum of the smallest values of each variable to the sum of the largest values of each variable. Consequently the sum of a pair of independent variates each on $(0,1)$ will lie in the interval $(0+0,1+1)$ (i.e. on $(0,2)$).

The shape of the density for the sum (as you'll find if you perform the convolution) is symmetric and triangular, though it's also obvious from direct inspection of a picture of the joint density:

enter image description here

The blue arrows show all the density at a fixed $x_1+x_2$; this is evaluated at each point along the red line. You can see the amount of density at each point increases linearly until the peak at 1, then decreases linearly again.


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