This wasn't made explicit in the video, but at 0:44 , those values are derives as follows: X = 0 E(Y|X=0) = .1 (this is the interception of y=1 and x=0) / (.4 + .1) (this is the marginal probability of x=0) = .2 X = 1 E(Y|X=1) = .4 (this is the interception of y=1 and x=1) / (.1 + .4) (this is the marginal probability of x=1) = .8
That is exactly the same question I had in my head. If x is 0 y is 0 (with a probability of .4) or 1 (with a probability of .1). How does this end up being an expected value of 0.2?
Revy Singh it is the correct answer from the definition of conditional expectation we have the suming over the y values for example for given x=0 we get 0*(0.4/0.5) + 1*(0.1/0.5) =0.2 ......I got the 0.5 from adding 0.4+0.1 as this give the probability of x=0
It’s fz(Z)={0.5 ,if z=0.2 or z=0.8. Remember Z is always E(Y|X) and those two (0.2, 0.8) are the only possible values of Z. So Z has a 50% chance of being one of the two at a time.
You get that result when you integrate the joint density function f(x,y) with respect to x, while setting the limits of the integration equal to the lower bound and upper bound limits of x (0 and 1). By doing this, you integrate the random variable X out of the joint density function, and you are left with the marginal density function (or the probability density function) of just the random variable Y.
if z=g(x) and g is monotone decreasing then F_Z(z)=P(Z = g^-1(z))= 1 - F_X(g^-1(z)) so f_Z(z)= -d/dz (F_X(g^-1(z))) = - f_X(g^-1(x)) d/dz g^-1(z) (main formula) here if Z = (3X+2)/(6X+3) =: g(X) then X = (-3+Z)/(6-3/Z) = g^-1(Z) in the last video he computed marginal of X from the joint x+y, it was f_X(x) = x+1/2 so insert it in the main formula -((-3+z)/(6-3/z) + 1/2) * ((-3+z)/(6-3/z))' = 1/(18 (-1 + 2 z)^3) check it here www.wolframalpha.com/input?i=-%28%28-3%2B2%2Fz%29%2F%286-3%2Fz%29%2B1%2F2%29*%28-1%2F%283+%281+-+2+z%29%5E2%29%29
This wasn't made explicit in the video, but at 0:44 , those values are derives as follows:
X = 0 E(Y|X=0) = .1 (this is the interception of y=1 and x=0) / (.4 + .1) (this is the marginal probability of x=0) = .2
X = 1 E(Y|X=1) = .4 (this is the interception of y=1 and x=1) / (.1 + .4) (this is the marginal probability of x=1) = .8
At 3:26 how do you get pdf of z
How did u get expected Value of y given x =0 equal to 0.2?
That is exactly the same question I had in my head. If x is 0 y is 0 (with a probability of .4) or 1 (with a probability of .1). How does this end up being an expected value of 0.2?
Revy Singh it is the correct answer from the definition of conditional expectation we have the suming over the y values for example for given x=0 we get 0*(0.4/0.5) + 1*(0.1/0.5) =0.2 ......I got the 0.5 from adding 0.4+0.1 as this give the probability of x=0
guys there is a sligt error
E(Y|x=x)=y*f(y|x)dy
Thank you, could you clarify how to get f(z)?
It’s fz(Z)={0.5 ,if z=0.2 or z=0.8. Remember Z is always E(Y|X) and those two (0.2, 0.8) are the only possible values of Z. So Z has a 50% chance of being one of the two at a time.
@@yaweli2968 I still don't understand where did he get 1/(18 (2 z - 1)^3) ? (timing i.e. 4:23)
as said at 8:15 how fy(y) = y+1/2
You get that result when you integrate the joint density function f(x,y) with respect to x, while setting the limits of the integration equal to the lower bound and upper bound limits of x (0 and 1). By doing this, you integrate the random variable X out of the joint density function, and you are left with the marginal density function (or the probability density function) of just the random variable Y.
if z=g(x) and g is monotone decreasing then F_Z(z)=P(Z = g^-1(z))= 1 - F_X(g^-1(z))
so f_Z(z)= -d/dz (F_X(g^-1(z))) = - f_X(g^-1(x)) d/dz g^-1(z) (main formula)
here if Z = (3X+2)/(6X+3) =: g(X) then X = (-3+Z)/(6-3/Z) = g^-1(Z)
in the last video he computed marginal of X from the joint x+y, it was f_X(x) = x+1/2
so insert it in the main formula
-((-3+z)/(6-3/z) + 1/2) * ((-3+z)/(6-3/z))' = 1/(18 (-1 + 2 z)^3)
check it here
www.wolframalpha.com/input?i=-%28%28-3%2B2%2Fz%29%2F%286-3%2Fz%29%2B1%2F2%29*%28-1%2F%283+%281+-+2+z%29%5E2%29%29