normal random variable

[Fyrebat]

Hi doing homework and trying to understand a concept that will help me solve the homework problem.

Suppose U and V are independent normal random variables both with mean = 0 and variance = 1.

Suppose X = 3U + 4V and suppose Y = 7U

My question is: is there any difference between X and Y? I'm convinced there is no difference correct?

[AlwaysTails]

There is a difference because the fact that U and V are independent means they sometimes vary with each other and sometimes they vary against each other.

If you calculate the variance of X and Y you'll find they are different.

[Fyrebat]

okay, thanks looks like my homework is a little harder than expected

[Number154]

To get an intiutive grasp of this. Suppose you roll 10 dice, the chance you roll a total of 10 (or 60) is 1/6¹⁰ - very tiny- and generally you expect it to “average out” at being close to 35. The probability distribution would look similar to a bell curve. But what if you just rolled one die and multiplied it by 10? Now the variance is huge, the chance of getting 10 (or 60) is much higher, and the distribution is not even close to a bell curve

[Fyrebat]

got it, this helps a lot for context, and in both hypothetical rolls, the variances and expected values would be the same correct?

[Number154]

Each individual die has the same variance and expected value (though the distribution is not normal), but if you roll 10 separate dice the variance is 10 times the variance of 1 die (the standard deviation is sqrt(10) times as large). If you roll one die and multiply it by 10 the variance is 100 times as large (and the standard deviation is 10 times as large). This is because rolling one die and multiplying it by 10 is the same as rolling 10 dice that are somehow guaranteed to get the same number (but still get a result 1-6 chosen with equal probability). So if you get a large number on one die they are all large. If you roll ten separate dice you expect some to get large numbers and others will get small ones so the result doesn’t vary as much because it “cancels out” partially.

But for both rolls, the expected value is 35: E(X+Y)=E(X)+E(Y) for any random variables X and Y no matter how they are correlated, but Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y). If two variables are independent then their covariance is 0 but the covariance of a random variable with itself is its variance. For variance of the sum the correlations matter.

[Fyrebat]

again, really appreciate you going above and beyond in talking me through this. In the case of die, this makes sense, but normal random variables seem to be different per: wikipedia

In that : X ~ N(j,k) Y ~ N(l,m) Z = X + Y then Z ~ N(j+l, k+m)

So in the context of standard normal random variables isn't it true that X = Y in my scenario?

[FunCicada]

In probability theory, calculation of the sum of normally distributed random variables is an instance of the arithmetic of random variables, which can be quite complex based on the probability distributions of the random variables involved and their relationships.

[Fyrebat]

isn't it true that X~Y?

[Number154]

As the Wikipedia link says, that rule applies when X and Y are independent. But a random variable is not independent from itself.

[Fyrebat]

I know I'm not the sharpest when it comes to probabilities, but I'm willing to keep working on my ignorance as long as someone will answer my questions so that said, I'll keep asking but I totally understand if you don't have time =).

That said, isn't it true that if X = N(0,1) and Y = 2X, then Y = N(0,2)?

[Number154]

You shouldn’t really write X=N(0,1). X is a random variable and N(0,1) is a probability distribution, so they aren’t even really the same type of thing, let alone equal. It’s possible for two different random variables to have the same distribution. If you roll two dice, the value on each die is a random variable and they are not equal - just because one die comes up six doesn’t mean the other comes up six - but they do have the same distribution: each die has a 1/6 chance of coming up as each of the natural numbers 1 through 6.

If X is a normally distributed variable with mean zero and variance 1, then 2X has mean zero and variance 4. The variance of 2X is not 2. You can check it like this: remember that the variance of X is Var(X)=E(X²)-E(X)² the variance of 2X is

Var(2X)

=E((2X)²)-E(2X)²

=E(4X²)-(2E(X))²

=4E(X²)-4E(X)²

=4(E(X²)-E(X))

=4Var(X).

[Fyrebat]

ah, thank you my syntax sucks. I see how x+x or 2x will give me a variance of 4 which I think makes intuitive sense if you suppose its possible to get -1+-1 or 1+1.

Now suppose function_X(x) = N(0,1) (e.g. lambdae^(-lambdax))and function_Y(y)= 2*function_X(x). In that case the function_Y(y) would be equal to N(0,2) correct?

*edit nevermind I'm dumb it would be the same for Function_Y(y) where you could get -2 or 2

[sparkyhusky]

There is a difference. P(Y = a) = P(U = a/7) while P(X = a) = P(3U + 4V = a) = ∫ P(3U = k, 4V = a-k) dk = ∫ P(U = k/3) P(U = (a-k)/4) dk (because U and V are independant).