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Find the joint CDF of the independent random variables $X$ and $Y$, where

$f_x(x)=x/2, 0\le x \le 2, $ and

$f_Y(y)=2y, 0 \le y \le 1$.

To do this, we can find the CDF separately for each of the marginal PDFs, and then multiply them together to get the joint CDF (since the variables are independent).

However, why doesn't it work to first multiply the marginal PDFs together, to get a joint PDF of $f_{X,Y}(x,y)=xy$, and then take the double integral (with respect to $x$ and then $y$) to get the joint CDF?

This is the way I thought to do it first but it seems like it wouldn't give the same answer; why not?

Thank you!

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If $X$ and $Y$ are independent random variables, then $$F_{X,Y}(x,y)=\int_{- \infty}^x \int_{- \infty}^y f_{X,Y}(w,v)dvdw = \int_{- \infty}^{x} \int_{- \infty}^{y} f_X(w)f_Y(v)dvdw$$ $$=\int_{- \infty}^{x} f_X(w)dw\int_{- \infty}^{y}f_Y(v)dv=F_X(x)F_Y(y).$$

Method 1 (joint pdf approach) gives: $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=xy,$$ if $0\leq x \leq 2, 0\leq y \leq 1$ and zero otherwise. Then $$F_{X,Y}(x,y)=\int_{0}^x \int_{0}^yf_{X,Y}(w,v)dvdw=\int_{0}^x \int_{0}^y wv dv dw=\frac{x^2y^2}{4}.$$

Method 2 (marginal cdf approach) gives: $$F_X(x) = \int_{0}^x \frac{w}{2}dw =\frac{x^2}{4}$$ $$F_Y(y) = \int_{0}^y 2vdv =y^2.$$ It follows $$F_{X,Y}(x,y)=F_X(x)F_Y(y)=\frac{x^2y^2}{4}.$$

I should note the cdf's are also piecewise functions: $F_X(x)$ and $F_Y(y)$ take a value of zero if $x \leq 0$ and $y \leq 0$, respectively, and take a value of 1 if $2 \leq x$ and $1 \leq y$, respectively. $F_{X,Y}(x,y)$ takes a value of zero if either $x \leq 0$ or $y \leq 0$, and takes a value of 1 if $2 \leq x$ and $1 \leq y$.

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  • $\begingroup$ Thank you so much!! This is super helpful. Why did you decide to set the bounds for the integral from $0$ to $x$ and $0$ to $y$, as opposed to $0$ to $2$ for $x$, and $0$ to $1$ for $y$, because aren't we given that those are the bounds? Thank you! $\endgroup$ – Sarina Mar 22 at 19:32
  • $\begingroup$ If we set the upper bounds to be 2 and 1, respectively, we do not get the cdf, we simply get 1, as we would be integrating over the entire support of both variables. $\endgroup$ – dlnB Mar 22 at 21:47

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