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Suppose that $X$ and $Y$ are independent random variables and $g$ is a real-valued function on $R$.

Show that $g(X)$and $Y$ are independent.

Okay, so i dont really know where to begin with this to be honest,

What i do know is that if $X$ and $Y$ are independent random variables then $$E(XY)+EXEY$$ and $$var(X+Y)=varX+varY$$

Would this information help me with my answer? Would i use one of the probability theorems for independence such as $P(A\cap B)=P(A)P(B)$?

EDIT: Would i be looking to find $$\mathbf{P}(g(X) = x, Y=y) = \mathbf{P}(g(X) = x) \cdot \mathbf{P}(Y=y),$$ to show independence?

If so what how would i go about this? Many thanks

Any help much appreciated, many thanks

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up vote 2 down vote accepted

Assume that $X$ and $Y$ are independent, that is $$ P(X\in A,Y\in B)=P(X\in A)P(Y\in B),\quad \text{for all}\; A,B\subseteq\mathbb{R}.\tag{1} $$ To show that $g(X)$ and $Y$ are independent, we have to show $(1)$ with $X$ replaced by $g(X)$.

Let $A,B\subseteq\mathbb{R}$. Since $g:\mathbb{R}\to\mathbb{R}$, we have that $g^{-1}(A)=\{x\in \mathbb{R}\mid g(x)\in A\}$ is also a subset of $\mathbb{R}$. Thus $$ \begin{align} P(g(X)\in A,Y\in B )&=P(X\in g^{-1}(A),Y\in B)=P(X\in g^{-1}(A))P(Y\in B)\\ &=P(g(X)\in A)P(Y\in B) \end{align} $$ and hence $g(X)$ and $Y$ are independent.

Note that in general we can only talk about the probability of $P(X\in A)$ for certain nice subsets $A$. These are called Borel sets, and hence $(1)$ is only required to hold for Borel sets $A$ and $B$. But this is probably out of the scope of your question.

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Thank you so much! This was amazing :) – Bernard.Mathews Oct 25 '13 at 4:43

If the joint density function of $X$ and $Y$ is the product of the individual densities then they are considered independent. Therefore, $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ the $X$ and $Y$ are independent. Let $Z=f(x)$. You have to show that:

$$f_{Z,Y}(z,y) = f_Z(z) f(y)$$

Start from the joint CDF of $Z$ and $Y$ and see if you can work it out.

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Thanks for your answer, Im not familiar with the notation you used though ($f_{X,Y}(x,y)=f_X(x)f_Y(y)$), what does $f_{X,Y}(x,y)$ mean? Sorry if this sounds stupid but i think my course may use different notation! I am in uk if that is relevent.....Thanks – Bernard.Mathews Oct 23 '13 at 3:46
The notation is used to indicate that the functions are potentially different. For example, $X$ can be a normal pdf and $Y$ can be a uniform pdf. In this case, $f_X$ refers to the normal and $f_Y$ refers to the uniform and $f_{X,Y}$ refers to the joint density function. Since, I am using the same symbol $f$, I need a way to communicate the idea that the pdfs of $X$, $Y$ and their joint all are potentially different functions. – response Oct 23 '13 at 3:51
Oh okay, i kind of understand, so do i have to show that $P(X=x, Y=y)=P(X=x)P(Y=y)$? – Bernard.Mathews Oct 23 '13 at 3:53
No, those probabilities would be $0$ as you have continuous random variables. Start from $P(Z \le z, Y \le y) = P(g(X) \le z, Y \le y)$. Then invert $g(X)$ to re-write the last probability, then use independence of $X$ and $Y$ and see if you can progress. – response Oct 23 '13 at 3:57
Ahh im completely confused!! This notation is not used in my lecture notes or model solutions! :/ – Bernard.Mathews Oct 23 '13 at 4:00

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