Chapter 6 Calculating Variance of Normal distribution

For calculating variance of sum of independent normal variables we need to add variance of normal distributions.
In Soln 6.30 pg-321 to find variance of 2X+Y

as variance of X and Y is shouldn't the variance of 2X+Y be 3 instead of 5.
Why are we doing square of 2 here?

Considering Exam Type Question on next page
variance of 3Y-4X is calculated by multiplying 3 and 4 to there respective variance and adding them.

 

I think it is worth you going back and revising the basics. Re-read sections 2.5 and 3.2 of that chapter.

 

I dont think these sections are applicable for normal distribution as mentioned on page 300.

N(X,s1^2)-N(Y,s2^2)~N(X+Y, s1^2 + s2^2).

So we must add the variance of these normal distributions to calculate variance of new normal distribution.

Please correct if I am not able to understand the concept.

 

Gaurav

I think you need to look at those sections. They work for any set of random variables including Normal.

In general, Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X, Y)
This result works for any set of two random variables X and Y.

Check yourself what happens when:
1. a = b = 1
2. a = 1, b = -1
3. X and Y are independent.

The only special additional result you should be aware of is when X and Y are Normal r.v, aX + bY is also Normal.

Now, if you have got this, re-read section 2.5 and 3.2.
After, try to solve the questions which are giving you trouble again.

 

Still not able to understand it.

 

Can you be bit more specific?

Which part of my earlier response did you not get?

 

Please check pg- 321 solution 6.30.
2X-Y~N(2*0-0, 2^2*1+(-1)^2*1)=N(0,5)

and on pg-322

Distribution of (Y1+Y2+Y3)-(X1+X2+X3+X4) which is:

(Y1 + Y2 + Y3) - ( X1 + X2+ X3 + X4) ~ N (3 * 1200 - 4 * 800, 3 * 300^2 + 4 * 100 )
~ N (400, 310000)

 

I believe your real issue is with computing variance.

I have actually given you the base formula to make you understand how this works. I will try again.

I have told you: Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X, Y)

For 6.30: try plugging in a = 2 and b = -1 and use the fact that X & Y are independent. You will get variance as 5.

For exam type question, do this:
1. Set W = X1+X2+X3+X4. Compute Var(W) applying the above rule repeatedly.
2. Set V = Y1+Y2+Y3. Compute Var(V) applying the above rule repeatedly.
3. Now you will have linear combination of V and W as (V - W) with a = 1 and b = -1. Compute Var(V - W)
(Note all r.v. involved are independent of each other)

You will get variance of 31,000 in the end.


As you can see that the base principle works for every case.

So, I would again recommend that you do what I suggested earlier. I have a feeling you need to get the basics understood first and then you can try the sums.