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Consider I am given two functions of one random variable each for example x=cos(at),y=rect(bt) where a and b are random variables.And I am given Probability density function for a and b then if I am asked if the two functions are independent or not so, I want to confirm that before proceeding I will have to convert the pdf of a and b to pdf of x and y or can I directly proceed with pdf of a and b by proceeding I mean checking if joint probability density is equal to product of marginal probability density.

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  • $\begingroup$ If $A$ and $B$ are independent random variables, then $g(A)$ and $h(B)$ also are independent; no need to check anything or do any pdf conversions etc. If $A$ and $B$ are not independent, then in general $g(A)$ and $h(B)$ are not independent either. $\endgroup$ – Dilip Sarwate Jul 5 at 11:30
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The canonical definition of independence of two random variables $X$ and $Y$ is

$X$ and $Y$ are called independent random variables if for every choice of Borel sets $B_1, B_2$, the events $\{X \in B_1\}$ and $\{Y \in B_2\}$ are independent events, that is, $$P\{X \in B_1, Y \in B_2\} = P\{X \in B_1\}P\{Y \in B_2\} \tag{1}$$

If you don't know what Borel sets are, rest assured that every set of real numbers you have encountered (and many more that you have never even dreamt of) is a Borel set. Choosing $B_1 = \{x\colon x \leq u\}$ and $B_2 = \{y\colon y \leq u\}$, Eq. $(1)$ tells us that $$P\{X\leq u, Y \leq v\} {=} P\{X\leq u\}P\{Y \leq v\}\tag{2}$$ which can also be expressed as $$F_{X,Y}(u,v) = F_X(u)F_Y(v).\tag{3}$$ It can be proved that if $(3)$ holds for all real numbers $u$ and $v$, then $(1)$ also holds and so $(3)$ is usually taken as the operational definition of independence of $X$ and $Y$:

$X$ and $Y$ are called independent random variables if $$F_{X,Y}(u,v) = F_X(u)F_Y(v) ~\text{for all}~u, v \in \mathbb R.\tag{4}$$


If $g(\cdot)$ and $h(\cdot)$ are real-valued (measurable) functions, then the random variables $W = g(X)$ and $Z = h(Y)$ are independent whenever $X$ and $Y$ are independent. This is because the events $\{W \leq u\}$ and $\{Z \leq v\}$ are the same as the events $\{X \in B_1\}$ and $\{Y \in B_2\}$ respectively where $B_1$ and $B_2$ are the pre-images of the sets $\{x\colon x \leq u\}$ and $\{y\colon y \leq v\}$ respectively under the maps $g(\cdot)$ and $h(\cdot)$. That is, \begin{align} B_1 &= \{w\colon g(w) \leq u\},\\ B_2 &= \{z\colon h(z) \leq v\}, \end{align} and from $(1)$, we know that $\{X \in B_1\}$ and $\{Y \in B_2\}$ are independent events. Thus, we have that $$P\{W \leq u, Z \leq v\} = F_{W,Z}(u,v) = P\{W \leq u\}P\{Z \leq v\} = F_{W}(u)F_{Z}(v)~\text{for all}~u, v \in \mathbb R $$ and so $g(X)$ and $h(Y)$ are also independent random variables. In summary,

Functions of independent random variables are independent random variables.

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  • $\begingroup$ In this case W(X,Y) and Q(X,Y) should also be independent isn't it. $\endgroup$ – Buzz bee Jul 5 at 18:27
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    $\begingroup$ @Buzzbee No, your notion is incorrect. See Fat32's answer. $\endgroup$ – Dilip Sarwate Jul 5 at 19:57
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    $\begingroup$ @Buzz bee: It's best to develop some inution for the concept of independence rather than just worrying about formulae. In the case of W(X,Y) and Q(X, Y), the random variables X and Y are in both functions, so clearly the value of W(X,Y) is going to related to the value of Q(X,Y) and vice versa. So, they ( W and Q) can't be independent because independence of two random variables ( W and Q can still be thought of as RV's ) says that knowing the value of Q doesn't provide any information about the value of the W and vice versa. Think of dependence as being related to providing information $\endgroup$ – mark leeds Jul 5 at 20:58
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    $\begingroup$ @markleeds +1 But while you are correct in pointing out that W and Q depend on both X and Y and so cannot be claimed to be independent, there are instances when for specific cases, independence does hold. For example, if $X$ and $Y$ are independent $N(0,\sigma^2)$ random variables, then $R=\sqrt{X^2+Y^2}$ and $\Theta=\arctan\left(\frac YX\right)$ are independent random variables even though both $R$ and $\Theta$ are functions of both $X$ and $Y$. I tell you this against my better judgment because BuzzBee will forget the caveats that this is a very special case and assert that ..... $\endgroup$ – Dilip Sarwate Jul 5 at 21:30
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    $\begingroup$ .... $g(X,Y)$ and $h(X,Y)$ are independent random variables for all choices of functions $g$ and $h$ and for all possible random variables $X$ and $Y$. Just watch for his next question on dsp.SE..... $\endgroup$ – Dilip Sarwate Jul 5 at 21:34
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It happens that if $X$ and $Y$ are independent then so will their functions $g(X)$ and $h(Y)$ be; but not $g(X,Y)$ and $h(X,Y)$.

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  • $\begingroup$ @Dilip Sarwate Plz can you mention how can I tell by inspection that where it went wrong after E[ZW]=E[Z]E[W] step in above answer. $\endgroup$ – Buzz bee Jul 5 at 11:57

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