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text: '$$\n\\newcommand{\\true}{\\text{True}}\n\\newcommand{\\false}{\\text{False}}\n\\newcommand{\\bP}{\\mathbb{P}} \n$$\n\n[summary: \n$$\n\\newcommand{\\true}{\\text{True}}\n\\newcommand{\\false}{\\text{False}}\n\\newcommand{\\bP}{\\mathbb{P}}\n$$\n\nSay $A$ and $B$ are independent [event_probability events], so $\\bP(A, B) = \\bP(A)\\bP(B).$ Then we can draw their joint probability distribution using the using the [496 square visualization] of probabilities:\n\n<img src="http://i.imgur.com/0off1db.png" width="312" height="272">\n\n]\n\n\n\nThis is what independence looks like, using the [496 square visualization] of probabilities:\n\n<img src="http://i.imgur.com/0off1db.png" width="390" height="338">\n\nWe can see that the [event_probability events] $A$ and $B$ don't interact; we say that $A$ and $B$ are *independent*. Whether we look at the whole square, or just the red part of\nthe square where $A$ is true, the probability of $B$ stays the same. In other words, $\\bP(B \\mid A) = \\bP(B)$. That's what we mean by independence: the\nprobability of $B$ doesn't change if you condition on $A$.\n\nOur square of probabilities can be generated by multiplying together the probability of $A$ and the probability of $B$:\n\n<img src="http://i.imgur.com/pjwcoTn.png" width="640" height="275">\n\nThis picture demonstrates another way to define what it means for $A$ and $B$ to be independent:\n\n$$\\bP(A, B) = \\bP(A)\\bP(B)\\ .$$\n\n\n\nIn terms of factoring a joint distribution\n--\n\nLet's contrast independence with non-independence. Here's a picture of two ordinary, non-independent events $A$ and $B$:\n\n<img src="http://i.imgur.com/6ZHSR0l.png" width="529" height="327">\n\n(If the meaning of this picture isn't clear, take a look at [496].)\n\nWe have the red blocks for $\\bP(A)$ and the blue blocks for $\\bP(\\neg A)$ lined up in columns. This means we've [factoring_probability factored] our\nprobability distribution using $A$ as the first factor: \n\n$$\\bP(A,B) = \\bP(A) \\bP(B \\mid A)\\ .$$\n\nWe could just as well have factored by $B$ first: $\\bP(A,B) = \\bP(B) \\bP( A \\mid B)\\ .$ Then we'd draw a picture like this:\n\n\n<img src="http://i.imgur.com/O0RNzxw.png" width="390" height="390">\n\n\n\n\nNow, here again is the picture of [4cf two independent events] $A$ and $B$:\n\n<img src="http://i.imgur.com/0off1db.png" width="390" height="338">\n\n\n\nIn this picture, there's red and blue lined-up columns for $\\bP(A)$ and $\\bP(\\neg A)$, and there's *also* dark and light lined-up rows for $\\bP(B)$ and\n$\\bP(\\neg B)$. It looks like we somehow [factoring_probability factored] our probability distribution $\\bP$ using both $A$ and \n$B$ as the first factor. \n\nIn fact, this is exactly what happened: since $A$ and $B$ are [4cf independent], we have that $\\bP(B \\mid A) = \\bP(B)$. So the diagram\nabove is actually factored according to $A$ first: $\\bP(A,B) = \\bP(A) \\bP(B \\mid A)$. It's just that $\\bP(B \\mid A)= \\bP(B) = \\bP(B \\mid \\neg A)$, since $B$\nis independent from $A$. So we don't need to have different ratios of dark to light (a.k.a. conditional probabilities of $B$) in the left and right columns:\n\n<img src="http://i.imgur.com/Nfiuz3d.png" width="618" height="420">\n\nIn this visualization, we can see what happens to the probability of $B$ when you condition on $A$ or on $\\neg A$: it doesn't change at all. The ratio of\n\\[the area where $B$ happens\\] to \\[the whole area\\], is the same as the ratio $\\bP(B \\mid A)$ where we only look at the area where $A$ happens, which is the\nsame as the ratio $\\bP(B \\mid \\neg A)$ where we only look at the area where $\\neg A$ happens. The fact that the probability of $B$ doesn't change when we\ncondition on $A$ is exactly what we mean when we say that $A$ and $B$ are independent.\n\nThe square diagram above is *also* factored according to $B$ first, using $\\bP(A,B) = \\bP(B) \\bP(A \\mid B)$. The red / blue ratios are the same in both rows\nbecause $\\bP(A \\mid B) = \\bP(A) = \\bP(A \\mid \\neg B)$, since $A$ and $B$ are independent:\n\n<img src="http://i.imgur.com/DfDljOL.png" width="636" height="468">\n\nWe couldn't do any of this stuff if the columns and rows didn't both line up. (Which is good, because then we'd have proved the false statement that any two\nevents are independent!)\n\nIn terms of multiplying marginal probabilities\n---\n\nAnother way to say that $A$ and $B$ are independent variables %note:We're using the [event_variable_equivalence equivalence] between [event_probability\nevents] and [binary_random_variable binary variables].% is that for any truth values $t_A,t_B \\in \\{\\true, \\false\\},$\n\n$$\\bP(A = t_A, B= t_B) = \\bP(A = t_A)\\bP(B = t_B)\\ .$$\n\n\n\nSo the [1rh joint probabilities] for $A$ and $B$ are computed by separately getting the probability of $A$ and the probability of $B$, and then\nmultiplying the two probabilities together. For example, say we want to compute the probability $\\bP(A, \\neg B) = \\bP(A = \\true, B = \\false)$. We start with\nthe [marginal_probability marginal probability] of $A$:\n\n<img src="http://i.imgur.com/ZnxqSMo.png" width="250" height="300">\n\nand the probability of $\\neg B$:\n\n<img src="http://i.imgur.com/txRlJyE.png" width="335" height="240">\n\nand then we multiply them:\n\n<img src="http://i.imgur.com/GOOnTuF.png" width="440" height="390">\n\n\nWe can get all the joint probabilities this way. So we can visualize the whole joint distribution as the thing that you get when you multiply two independent\nprobability distributions together. We just overlay the two distributions: \n\n<img src="http://i.imgur.com/X4FSciB.png" width="532" height="816">\n\nTo be a little more mathematically elegant, we'd use the [topological_product topological product of two spaces] shown earlier to draw the joint distribution\nas a product of the distributions of $A$ and $B$: \n\n<img src="http://i.imgur.com/pjwcoTn.png" width="640" height="275">\n\n\n\n',
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