sum of normal distribution help

#1
X~N(3,5) Y~N(-7,2)
What is the distribution of Z=4X-Y/3 ?

is it just

=- Y/3 + 4X
=- Y~N(-7/3,2/3) + X~N(-7/3 + 3,2/3 + 5)
= X~N(-7/3 + 3,2/3 + 5) - Y~N(-7/3,2/3)

?

What is the covariance of X and Z?

I don't understand how to do this...
 

BGM

TS Contributor
#2
Again you need several basic properties:

1. If \( X, Y\) are independent normal random variables,
then \( Z = X + Y \) is also normally distributed.

2. If \( X \) is a normal random variable,
then \( aX + b \) is also normally distributed.

3. \( E[X + Y] = E[X] + E[Y] \)

4. \( Var[X + Y] = Var[X] + Var[Y] + 2Cov[X, Y]\)

5. If \( X, Y\) are independent,
then \( Cov[X, Y] = 0 \)

6. \( Cov[X + Y, Z] = Cov[X, Z] + Cov[Y, Z] \)
 
#3
oh right, i forgot to mention that X, and Y are independent.

i'm not understanding how to multiply normal distributions by constants,
is 4*X~N(3,5) = X~N(12,20) ?

EDIT: i now get Z~N(43/3, 20.011) is this right?

and i think i figured out the covariance

Cov(X,Z)=Coz(X,4X-1/3*Y)
=Coz(X+0Y,4X-1/3*Y)
=4Cov(X,X)-1/3 Cov(X,Y)
=4Var(X)-1/3(0) <--- by independence
=4Var(X)
=4*5
=20

Is this right?
 
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