Think of each value of Y as a scenario. Under each scenario X can take value x with some probability. To find marginal probability, apply total law probability.
No,you're not running the summation twice , you are running it once.Summing over PAIRS (x,y) such that g(x,y) = z. You get a single value z each time , and u sum over those z's. Essentially what i mean to say is that u sum over the OUTPUT of g(x,y) , not over the INPUT of g(x,y) which is X and Y , and then you would indeed need double summation. An example , lets say g(x,y)= x+y , then for the pair (1,2) u have z=3.You add the 3 to your sum , not the 1 or the 2 to some double sum
@@DeadPool-jt1ci We could do it with double summation too. Like For each value of x consider all value of y such that g(x,y) = z. This wont sum twice. Like its being done in Expectation.
The explanation of Dead Pool is not quite correct. You indeed sum over pairs (x,y) that give the same particular z, but do not sum over z's, since z is the variable at the left hand side and can not be the index of summation.
Think of each value of Y as a scenario. Under each scenario X can take value x with some probability. To find marginal probability, apply total law probability.
excellent lecture mr
At 2:17, could someone explain to me why both P(Y=2) and P(X=1) are clearly greater than 0? But P(X=1 and Y = 2) is 0? Thank you!
why this cannot be true? Y can equal 2 and X can equal 1 but they cannot both equal to that number simultaneously
in this case they are not independent
at 9:18 shouldn't it be double summation to be more accurate?
No,you're not running the summation twice , you are running it once.Summing over PAIRS (x,y) such that g(x,y) = z. You get a single value z each time , and u sum over those z's. Essentially what i mean to say is that u sum over the OUTPUT of g(x,y) , not over the INPUT of g(x,y) which is X and Y , and then you would indeed need double summation.
An example , lets say g(x,y)= x+y , then for the pair (1,2) u have z=3.You add the 3 to your sum , not the 1 or the 2 to some double sum
@@DeadPool-jt1ci got it. Thanks
@@DeadPool-jt1ci We could do it with double summation too. Like For each value of x consider all value of y such that g(x,y) = z. This wont sum twice. Like its being done in Expectation.
The explanation of Dead Pool is not quite correct. You indeed sum over pairs (x,y) that give the same particular z, but do not sum over z's, since z is the variable at the left hand side and can not be the index of summation.
Very helpful!
awesome
sadly, I can't understand very much. Why don't you demonstrate some specific example with real figure ?
👍👍👍👍👍👍👍👍
3:15
the little russian accent made me sure that math and probability were made by russians lol ( foreigner student in russia )
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Thank you Kolmogorov! (Prof. Tsitsiklis is Greek btw)
2min
it is not clear explained .