In: Math
1) The PDF of a Gaussian random variable is given by fx(x).
fx(x)= (1/(3*sqrt(2pi) )*e^((x-4)^2)/18
determine
a.) P(X > 4) b). P(X > 0). c). P(X < -2).
2) The joint PDF of random variables X and Y is given by
fxy(x,y)=Ke^-(x+y), x>0 , y>0
Determine
a. The constant k.
b. The marginal PDF fX(x).
c. The marginal PDF fY(y).
d. The conditional PDF fX|Y(x|y). Note
fX|Y(x|y) =
fxy(x,y)/fY(y)
e. Are X and Y independent.
2)
Note that for a function to be a pdf, the cumulative distribution in the given domain should sum up to 1. ( i.e. the volume under the given surface function should be = 1)
The last part can also be done using intuition from part d.
We see that the conditional pdf x|y is same as pdf of x. This means that the occurrence of x doesn't depend on y. Therefore they are independent.