The value of the expression :
is equal to = ?
Answers
Step-by-step explanation:
Problem
Let X,Y and Z be three jointly continuous random variables with joint PDF
fXYZ(x,y,z)=⎧⎩⎨⎪⎪13(x+2y+3z)00≤x,y,z≤1otherwise
Find the joint PDF of X and Y, fXY(x,y).
Solution
Problem
Let X,Y and Z be three independent random variables with X∼N(μ,σ2), and Y,Z∼Uniform(0,2). We also know that
E[X2Y+XYZ]=13,E[XY2+ZX2]=14.
Find μ and σ.
Solution
Problem
Let X1, X2, and X3 be three i.i.d Bernoulli(p) random variables and
Y1=max(X1,X2),Y2=max(X1,X3),Y3=max(X2,X3),Y=Y1+Y2+Y3.
Find EY and Var(Y).
Solution
Problem
Let MX(s) be finite for s∈[−c,c], where c>0. Show that MGF of Y=aX+b is given by
MY(s)=esbMX(as),
and it is finite in [−c|a|,c|a|].
Solution
Problem
Let Z∼N(0,1) Find the MGF of Z. Extend your result to X∼N(μ,σ).
Solution
Problem
Let Y=X1+X2+X3+...+Xn, where Xi's are independent and Xi∼Poisson(λi). Find the distribution of Y.
Solution
Problem
Probability Generating Functions (PGFs): For many important discrete random variables, the range is a subset of {0,1,2,...}. For these random variables it is usually more useful to work with probability generating functions (PGF)s defined as
GX(z)=E[zX]=∑n=0∞P(X=n)zn,
for all z∈R that GX(z) is finite.
Show that GX(z) is always finite for |z|≤1.
Show that if X and Y are independent, then
GX+Y(z)=GX(z)GY(z).
Show that
1k!dkGX(z)dzk|z=0=P(X=k).
Show that
dkGX(z)dzk|z=1=E[X(X−1)(X−2)...(X−k+1)].
Solution
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