Re: how to determine the variance of dependant variable when all the var. of ind. variables r known in a linear relation

Stephen J. Herschkorn wrote:
Greg Heath wrote:
Stephen J. Herschkorn wrote:
Greg Heath wrote:
Stephen J. Herschkorn wrote:
Greg Heath wrote:
Jerry wrote:
hi, here is the question,

if we know,
y = a * X1 + b * X2 + c

and we also know the variance of X1 and X2, how to evaluate the
variance of the Y?

Plug in and simplify:

VAR(y) = E{ (y - E{y} )^2 }

Hope this helps.

Not much. It is much simpler than that.

Of course it is. However, as a self-proclaimed tutor, don't you think
Jerry will benefit more by working it out himself?

Not at all.

I think Jerry should be the judge of that:

HEY JERRY!

Did you learn more by working it out yourself?

The OP should be familiar with basic facts about variance
and know when and how to apply them.

And if he isn't, would merely stating the answer really be of
any meaningful help to him?

In college I was educated with the British/Australian
philosophy that, during learning, deducing the correct
approach from first principles is more important than
memorization and/or getting the correct answer.

I am forever grateful that most of my Engineering,
Applied Math and Applied Physics professors believed
in that approach (Whereas most of my Math, Chemistry
and Physics professors did not).

When I taught Engineering and supervised Researchers
I continued that philosophy.

Now that I'm retired I see no reason to change,

Our styles of tutoring appears to be orthogonal.

I prefer to teach a man how to fish.

A student of probability should learn the basic facts that Var(a X) =
a^2 Var(X) and, when X and Y are independent, Var(X + Y) = Var(X)
+ Var(Y).

Those are derived results. Not basic. Who said Jerry's X1 and X2
were independent?

Maybe Jerry is not a probability student. May he is and but hasn't
internalized that first fact yet...

In the OP's case, proceding from the definition of variance
is inappropriate.

Difference in opinion.

HEY JERRY!

What do you think?

Would you have someone compute integral(x=0..1,
x^2) by working directly from the definition of the Riemann integral?

Irrelevant.

Computations and derivations are two different animals.

What would you say if some one asked you what the derivative
of x^4 was? 4*x^3?

In some circumstances I would say deduce the general answer
for x^n by applying the definition

lim(dx-->0){ (f(x+dx)-f(x))/dx}

to x, x^2 and x^3.

Teach a man how to fish.

You yourself said you are not a statistician (or probabilist, I take
it). This is not the first time you have offered misleading suggestions
in this group.

And definitely not the last. However, I don't consider this
as one of times. What was so misleading about it?

Of course I assume Jerry took a good look at his answer
and learned something from it.

That's fine, but please don't disparage others' reputations

Touchy, touchy.

Why do you feel that your reputation is disparaged?

Because we have a difference of opinion in tutoring styles?

Please.

when it is pointed out that these suggestions are not very
helpful.

I welcome criticism. Especially when I can learn from it.

The only thing I learned from this exchange is that ...

Well, I'll stop here.

Have a good day.

WAIT!

How would you have replied to Jerry's question?

Greg

.

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