Let X be a standard normal random variable. Let Y be a continuous random variable such that
fY|X(y|x)=1/√2π−−exp(−(y+2x)^2/2).
Find E[Y|X=x] (as a function of x, in standard notation) and E[Y].
E[Y|X=x]=
unanswered
E[Y]= unanswered
Compute Cov(X,Y).
Cov(X,Y)= unanswered
The conditional PDF of X given Y=y is of the form
α(y)exp{−quadratic(x,y)}
By examining the coefficients of the quadratic function in the exponent, find E[X∣Y=y] and Var(X∣Y=y).
E[X∣Y=y]=
unanswered
Var(X∣Y=y)=
E[Y|X=x] = -2x
E[Y] = 0
Cov(X,Y)= -2
E[X∣Y=y]=-y/2
Var(X∣Y=y)=1/16
Cov(X,Y)= -2
E[X∣Y=y]= -2/5*y
Var(X∣Y=y)= 1/5
E[Y|X=x] = -2x
E[Y] = 0
Cov(X,Y)= -2
E[X∣Y=y]=-2y/5
Var(X∣Y=y)=1/5
why the cov(X,Y) = -2
So if we look at the formula, fX|Y(x|y) ~ N(-2x,1), so
E[Y|X=x] = -2x
E[Y] = 0
what kind of math is this😵😵😵😵😵🤯🤯🤯🤯🤯
To find E[Y|X=x], we need to calculate the conditional expected value of Y given a specific value of X. In this case, we are given the conditional PDF of Y|X, which is a normal distribution with mean -2x and standard deviation 1.
The formula for the conditional expected value is given by:
E[Y|X=x] = ∫y*fY|X(y|x) dy
Substituting the given PDF, we have:
E[Y|X=x] = ∫y*(1/√(2π))*exp(-(y+2x)^2/2) dy
To find the integral, we can use techniques like completing the square or using a standard normal table. However, since this is a complex integral, it is not easily solvable analytically. In practice, numerical methods like integration techniques or computer software are used to approximate the integral.
Similarly, to find E[Y], we need to integrate the entire joint PDF of X and Y:
E[Y] = ∫∫y*fX,Y(x, y) dx dy
However, the joint PDF fX,Y(x, y) is not given in the question, so we cannot calculate E[Y] without additional information.
Moving on to Cov(X, Y), the covariance between two random variables X and Y is defined as:
Cov(X, Y) = E[(X - E[X])(Y - E[Y])]
To calculate this, we need to find E[X], E[Y], E[XY], and plug them into the formula above. Since we do not have the information required to calculate E[Y] and E[XY], we cannot compute Cov(X, Y) at this time.
Finally, let's consider the conditional PDF of X given Y=y. According to the given form, it has the structure:
α(y)exp{-quadratic(x,y)}
By examining the coefficients of the quadratic function in the exponent, we can determine the conditional mean (E[X|Y=y]) and variance (Var(X|Y=y)). However, since the specific form of the quadratic function is not provided in the question, we cannot determine E[X|Y=y] and Var(X|Y=y) without more information.
In summary, without additional information or specific forms of the PDFs, we cannot find E[Y|X=x], E[Y], Cov(X,Y), E[X|Y=y), and Var(X|Y=y).
E[Y|X=x] = -2x
E[Y] = 0
Cov(X,Y)= -2
E[X∣Y=y]=-y/2
Var(X∣Y=y)=1/4
The rest I'm not sure:
cov(X,Y) = 0
E[X|Y=y] = -y/2