Thread: unexpected expectation

For a random variable X, find E[Y] where Y=(X-E[X])/(standard deviation of X).

I've only tinkered with this a few minutes, and am working on it now but am not seeing it as terribly obvious. I'm just looking for a "Dude, its pretty straightforward, just keep tinkering" or a "Well, it ain't bad but without *this* or *that* identity, you're going to go cross-eyed staring at your page."

I just don't want to spend an hour on what may be a 3-minute algebra march.

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I found that expression used several times in a few different spots in my text. It seems like an important proportion, but there is no discussion nor even explicit mention of it solo. Maybe if someone could just explain what that proportion is briefly, the lights might come on for me...

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It zero, isn't it? If I think about Z=X-E[X], the mean (expectation) of Z is zero (this much is quite straightforward). The only difference between my Z and the Y in the problem statement is that Y is scaled by the standard deviation. I think E[Y]=0. Still, feel free to confirm this or shoot it down like a lame duck.

Hello,

Haha I like your thread's title

It zero, isn't it? If I think about Z=X-E[X], the mean (expectation) of Z is zero (this much is quite straightforward). The only difference between my Z and the Y in the problem statement is that Y is scaled by the standard deviation. I think E[Y]=0. Still, feel free to confirm this or shoot it down like a lame duck.

Yes it's 0 There's no worry to have about that.

But where the standard deviation of X intervenes it's in the variance of Y.
Indeed, recall that $\displaystyle V(X+a)=V(X)$, where a is a constant, and that $\displaystyle V(aX)=a^2V(X)$, again where a is a constant.

So here, we'd have $\displaystyle V(Y)=1$

I found that expression used several times in a few different spots in my text. It seems like an important proportion, but there is no discussion nor even explicit mention of it solo. Maybe if someone could just explain what that proportion is briefly, the lights might come on for me...

It helps transforming a normal distribution into a standard normal distribution (have a look here : Normal distribution - Wikipedia, the free encyclopedia ). It's useful because we know the centiles of the standard normal distribution (there are tables called "z-tables"), but there is no table giving directly the centiles of a normal distribution in general.

Enjoy !