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We know that the joint probability function of two independent random variables is just the product of their respective pdfs. On the same lines, .can we say that if we multiply the cumulative density functions(CDFs) of those two random variables, the resulting function will be the CDF of joint distribution? Like f1 and f2 are the two pdfs of the independent random variable. The joint pdf will simply be f1f2. If F1 and F2 are their CDFs.Can we say that F=F1F2 represents their joint CDF?

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    $\begingroup$ Short answer: yes. $\endgroup$ Commented Jun 12, 2020 at 21:49
  • $\begingroup$ Thanks Adrian.Can you please suggest me any text regarding this concept which I can see. $\endgroup$
    – Syed yunus
    Commented Jun 12, 2020 at 22:02
  • $\begingroup$ any book or any other source $\endgroup$
    – Syed yunus
    Commented Jun 12, 2020 at 22:03

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Yeah. Its just a simple exercise: Suppose $X$ is a random variable distributed according to pdf $g$ and cdf $G$. Similarly $Y$ is a random variable distributed according to pdf $h$ and pdf $H$. Let $f$ be the pdf of the joint distribution of $(X,Y)$.

$X$ and $Y$ are independent: i.e. $f(x,y)=g(x)h(y)$ for all $(x,y)\in X\times Y$.

Then we have the following: \begin{align} F(x,y) &= Pr(X\leq x, Y\leq y)\\ & = \int_{X\leq x}\int_{Y\leq y}f(s,t)dtds\\ & = \int_{X\leq x}\int_{Y\leq y}g(s)h(t)dtds\\ & = \int_{X\leq x}g(s)H(y)ds\\ & = G(x)H(y) \end{align}

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  • $\begingroup$ Thank You so much Shomak for explaining so nicely and clearly with proof. $\endgroup$
    – Syed yunus
    Commented Jun 12, 2020 at 22:14
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    $\begingroup$ No problem at all. The second last equality secretly employs Fubini's theorem, in case anyone is wondering. $\endgroup$
    – user288073
    Commented Jun 12, 2020 at 22:20

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