# What does addition operation on RVs mean?

• Nov 6th 2011, 08:30 PM
rebghb
What does addition operation on RVs mean?
Hello Everyone!

I've read in a text somewhere what looks like the following:

X and Y are two RVs. Let Y = aX + b for some two real a and b.

First of all, what does scalar multiplication of an RV mean?

Second, what does addition of a constant to an RV mean?

Any help is appreciated!
• Nov 7th 2011, 01:20 AM
Moo
Re: What does addition operation on RVs mean?
Hello,

Basically, the RV takes some values. Suppose you have X following a Bernoulli distribution (the most simple) with parameter p.
Then P(X=0)=1-p and P(X=1)=p.
What is the distribution of Y=aX+b ? Its only possible values will depend on those of X : a*0+b=b and a*1+b=a+b.
So P(Y=b)=P(a*X+b=b)=P(X=0)=1-p and P(Y=a+b)=p.
• Nov 7th 2011, 02:25 AM
rebghb
Re: What does addition operation on RVs mean?
Alright yeah I was always thinking that an RV, for instance X, is only an abstraction.

Thanks!
• Nov 7th 2011, 03:53 AM
Moo
Re: What does addition operation on RVs mean?
The real definition of a rv is that it's a function that goes from $\displaystyle \Omega$ (the set of all events) and $\displaystyle \mathbb R$ for example (it could be $\displaystyle \mathbb C$ as well).

When one actually write $\displaystyle X$, it's $\displaystyle X(\omega), \omega \in\Omega$.
When you write a*f(x)+b it's the same.

A probability P measures the 'quantity' of $\displaystyle \omega$ such that the event is realised : $\displaystyle P(X=5414775)=P(\omega ~:~ X(\omega)=5414775)$. Here, the event is {X=5414775}.

I'm trying to give you a deeper explanation, if you don't understand, don't worry you'll know it in time if you continue studying probability :)
• Nov 7th 2011, 07:09 AM
rebghb
Re: What does addition operation on RVs mean?
Well I'm studying about random processes right now :D

You know it's always technical around here, very mechanic. I know my way around RVs, integrals, etc. but I only started to learn how they popped up a while after I first got to learn them.

Thanks again :)