ChaseB2039
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Do we need to know the variance formulas etc? I've never seen it in a hsc paper and my teacher said we don't.
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Variance (1 Viewer)
- Thread starter ChaseB2039
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If you are talking about the PDF function stuff, it’s probably advised that you should just in case bc I’ve seen a lot of E(X) and stuff around that. But it’s not too hard bc it’s just x^2 multiplied by ur given function minus the E(X)^2
ExtremelyBoredUser
Bored Uni Student
Generally and without prior knowledge,
 = \left\{ \begin{array}{lr} \int_{-\infty}^{\infty} x^2 f_{X}(x) \, dx - \left(\int_{-\infty}^{\infty} x f_{X}(x) \, dx\right)^2 & : X \text{ is continuous}\\ \sum_{\text{all }x} x^2p_{X}(x) - \left(\sum_{\text{all }x} xp_{X}(x)\right)^2 & : X \text{ is discrete} \end{array} \right. \end{displaymath} )
however if you know)
already, then you can use the normal formulas.
and I don't think you'll need to know anything beyond this but if it helps, you can use LoTUS
] = \int_{-\infty}^{\infty} g(x)f_{X}(x) \, dx)
which will help for more messier/harder questions but I honestly doubt they'll put anything beyond this.
Just think of the computations for the Variance/Expected Value as sums and depending on the R.V X, you either compute it through integrals or normal sums.
however if you know
and I don't think you'll need to know anything beyond this but if it helps, you can use LoTUS
which will help for more messier/harder questions but I honestly doubt they'll put anything beyond this.
Just think of the computations for the Variance/Expected Value as sums and depending on the R.V X, you either compute it through integrals or normal sums.
ughh hate this topic so muchGenerally and without prior knowledge,
however if you know
and I don't think you'll need to know anything beyond this but if it helps, you can use LoTUS
which will help for more messier/harder questions but I honestly doubt they'll put anything beyond this.
Just think of the computations for the Variance/Expected Value as sums and depending on the R.V X, you either compute it through integrals or normal sums.
Average Boreduser
Rising Renewal
it aint that bad tho be fr. conceptually might be annoying to visualise given most proofs in that topic is out of syllabus but perms and combs is 10x as annoying lmao
liamkk112
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just visualise the discrete version, the continuous version follows from there
for example, for a discrete variable we have E[X]= sum over i of x_i P(x_i), or in other words for each event in the sample space we sum over the magnitude of each event times the probability of the event occurring. so really this is just the ”weighted average” of a random variable (if you want consider when there’s n events, and each event is equally likely so P(x_i) = 1/n for any i; then E[X] = (x_1 + x_2+…+x_n)/n, which is just the usual average you might be used to).
then if you think of an integral as the continuous version of a sum, all we are doing is changing the sum to the integral for a continuous random variable; so E[X]=integral of x p(x) dx from -inf to +inf; we integrate from -inf to +inf because we want to consider all of the events
ExtremelyBoredUser
Bored Uni Student
Integrals can be treated effectively just sums so when you look it from that angle it's no real surprise why the discrete/cts version is identical ignoring the sum notation... you'll learn this in first year maths when you study about Riemann integration. Regardless, think more about the nature of the variable rather than the formula, as liam pointed out, as it'll make sense how you get the expected value/variance and other moments from there.