# Sum of random variable

Could we think about joint distribution as sum of random variables? Is the sum of random variables is the same thing as joint distribution of this variables? Would be appreciated!

• No. Have you tried reading an elementary probability book? Think about the special case of X and Y being deterministic. If you know X and Y, then you know X+Y. But if you know only X+Y, you don't know what X and Y are individually (or jointly). Jul 12, 2016 at 13:17
• I am reminded of the Mad Hatter's riddle, "How is a raven like a writing desk?"
– whuber
Jul 12, 2016 at 13:23
• In the first page (47) you can see that for calculation of mean value used a joint distribution of random variables, so why? ocw.mit.edu/courses/physics/… Jul 12, 2016 at 13:57
• Your question and your comments are a little baffling: If you want to compute any property of a pair of random variables, your calculation ultimately depends on the distribution of those variables. But the property that you calculate--whether it be their sum, their means, or literally anything else--obviously is not the same thing as their distribution.
– whuber
Jul 12, 2016 at 14:19

Sum of Independent random variables is called convolution of probability distributions.

As others have pointed out, these two are entirely different things. To understand why, consider two independent standard normal distributions, $X$ and $Y$. Their joint distribution is a two dimensional gaussian with mean $(0,0)$ and covariance matrix $I_2$. But the sum of the two random variables will be a 1 dimensional gaussian with mean $E(X) + E(Y)$ and variance $Var(X)+Var(Y)$.

I think Mark L. Stone has a good answer. The answer is definitely NO.

I will give you another toy example to highlight the differences. Think about you have $2$ discrete binary random variables $X$ and $Y$, i.e., each of them can take $0$ or $1$ with different probability, for each random variable the probability mass function can be described with a $1 \times 2$ table.

• The sum is still a binary random variable but can take from $0$ to $2$, the probability mass function can be described it with a $1 \times 3$ table.
• The join distribution is not a binary random variable, but a $2 \times 2$ table.

EDIT: To answer your question in the comment, I would extend the binary discrete example.

Say $X$ and $Y$ are iid, and

$P(X=0)=0.2$, $P(X=1)=0.8$,

$P(Y=0)=0.2$, $P(Y=1)=0.8$.

Let $Z=X+Y$

Then $P(Z=0)=0.2*0.2=0.04$ ,

$P(Z=2)=0.8*0.8=0.64$,

$P(Z=1)=0.2*0.8+0.8*0.2=0.32$.

The mean of $Z$ is

$0*0.04+2*0.64+1*0.32=0.16$

• ocw.mit.edu/courses/physics/… In the first page (47) you can see that for calculation of mean value used a joint distribution of random variables, so why? Jul 12, 2016 at 13:58
• The lecture notes you mentioned is trying to calculate the mean for one random variable that comes from sum of other iid random variables. The reason you see joint distribution there is because it is the definition of the mean. Jul 12, 2016 at 14:37
• The derived random variable comes from many other iid random variables. To calculate the mean you need to integrate over the joint / all the possible configurations on all iid variables. Jul 12, 2016 at 14:49
• Sorry your math is really unclear for me, and do not know your question. Jul 12, 2016 at 15:34
• Sorry, but i didn't understand your example! Is it written correctly? We have two discrete binary random variable $X$ and $Y$, and $Z=X+Y$.The convolution of two independent identically distributed Bernoulli random variables is a Binomial random variable, so $Z$ is Bernoulli random variable, but in your example $Z$ has three possible outcomes {0,1,2}, but instead it should have two outcomes. Am I right? Jul 12, 2016 at 16:47