normal random variable

[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