# Independent distributions

I have a confusion regarding the notion of independence of distributions. what is meant by saying two distributions are independent..? suppose I have two normal distributions with means 1,-1 and variances 1/2, 1/2 respectively. then is the sum of the pdf is also a pdf of normal distribution or not..?

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Are you talking about the sum of two random variables or the sum of two probability density functions? They're quite different things. – Eckhard Jan 10 '13 at 14:49
Why the downvote? It's the first question of OP, and totally not the worst among the first questions I've seen. – Ilya Jan 10 '13 at 15:31
To complement @Stefan's answer, note that one never says that two distributions are independent (or not), one says that random variables (with given distributions) are (or are not) independent. (In fact, distributions could be independent if they were themselves random, but this notably more sophisticated context is clearly not the one the OP has in mind.) – Did Jan 12 '13 at 9:57

Let $X\sim\mathcal{N}(1,\frac{1}{2})$ and $Y\sim\mathcal{N}(-1,\frac{1}{2})$ be two normal distributed random variables with mean $1$ and $-1$ respectively, both having variance $\frac{1}{2}$. We haven't said anything about the relationship between the two variables yet, and unless we do so, we can not say anything about $X+Y$ for sure. Let us therefore introduce independency:
We say that $X$ and $Y$ are independent if $$P(X\in A,X\in B)=P(X\in A)P(X\in B),\quad \text{for all }\,A,B\in\mathcal{B}(\mathbb{R}),$$ which is equivalent to saying that $$P(X\leq x,Y\leq y)=P(X\leq x)P(Y\leq y),\quad x,y\in\mathbb{R},$$ i.e. the joint distribution function equals the product of the two marginal distribution functions. If $X$ and $Y$ are independent, then $X+Y$ is again normal distributed (this is a special property of the normal distribution) with mean $$E[X+Y]=E[X]+E[Y]=1+(-1)=0,$$ and variance $$\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)=1.$$