Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have two random variables $X_1$ and $X_2$ with $P(X_1=x_1)=(1-p)^{1-x_1}p^{x_1}$ and for $X_2$ the same and $x_1 \in \{0,1\}$. I want to know how $Y=X_1+X_2$ is distributed. This is what i did: $P(Y=y)=P(X_1+X_2=y)=\sum_{x=0}^1 P(X_1=x, X_2=y-x)=\sum_{x=0}^1 P(X_1=x)P( X_2=y-x)=$ $\sum_{x=0}^1 (1-p)^{1-x}p^x(1-p)^{1-(y-x)}p^{y-x}=\sum_{x=0}^1 (1-p)^{2-y}p^y=2p^y(1-p)^{2-y}$

I think this is wrong because I think the answer should be ${2\choose y} p^y(1-p)^{2-y}$ So what did I do wrong? $X_1$ and $X_2$ are independent.

share|improve this question
    
There's not enough information to find the distribution of $Y$; you need to know the joint distribution of $X_1$ and $X_2$. Perhaps you intended them to be independent? Also, what's $k$ doing there? It never occurs again after you introduce it. –  joriki Feb 1 '13 at 10:42
    
@joriki I meant $x_1$ and yes, they are independent. –  Badshah Feb 1 '13 at 10:46
    
Please add that missing information to the question itself; people shouldn't have to read the comments in order to make sense of the question. –  joriki Feb 1 '13 at 10:48

2 Answers 2

up vote 3 down vote accepted

This looks far too complicated.

You have two random variables, each of which takes - the value $1$ with probability $p$ and - the value $0$ with probability $1-p$.

If they are independent then their sum takes

  • the value $2$ with probability $p^2$
  • the value $1$ with probability $2p(1-p)$
  • the value $0$ with probability $(1-p)^2$

The sum is a binomial random variable.

A particular problem with your expressions is that you have not restricted the values $X_2$ can take. For example when $y=0$ the expression $\sum_{x=0}^1 P(X_1 =x)P(X_2 =y−x)$ means $P(X_1 =0)P(X_2 =0−0) + P(X_1 =1)P(X_2 =0-1)$ but you should not have the second term in that sum since $P(X_2 =-1) =0$ rather than the positive value you give it

share|improve this answer
    
so which restriction should I give to $X_2$? –  Badshah Feb 1 '13 at 10:58
    
@Badshah: It has to be $0$ or $1$ so you need $$\sum_{x=\max(0,y-1)}^{\min(1,y)} P(X_1 =x)P(X_2 =y−x)$$ –  Henry Feb 1 '13 at 11:08

What you wrote is not correct (first step) because according to that $P(Y=0) = \sum_{x=0}^1 P(X_1=x,X_2=-x)$ and $X_2$ can't be negative. Moreover, if $X_1=x$ and $X_2=y-x$ they can't be independent.

Sorry guys but I don't see where I can put this as a comment :S :S

share|improve this answer
    
why cant they be independent? –  Badshah Feb 1 '13 at 11:03
    
Because as you write them, one depends on the result of the other. –  dann Feb 1 '13 at 11:08
    
I would use either a reasoning based on "observing", draw a Venn's diagram and you'll see easily what is the distribution of $Y$. Or mathematically, the sum of two indep. r.v. is the convolution. –  dann Feb 1 '13 at 11:12
    
Mmm sorry you're right, I see it now. They are independent since $X_1$ and $X_2$ are ssumed to be ind. –  dann Feb 1 '13 at 11:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.