How Can You Calculate the Expected Value E(X^2(3Y-1))?

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  • #1
adamwitt
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Expected Values... E(X^2(3Y-1))

Homework Statement



What is the Expected Value of E(X2(3Y-1))

Homework Equations



Properties of Expected Values.
E(X+c) = E(X) + c
E(X+Y) = E(X) + E(Y)


The Attempt at a Solution



I tried googling for a few of the properties I could use but I'm not sure which ones to use, or maybe I haven't found the required property?
I have been given the data and worked out the marginal probability mass functions so I can work out E(X), E(Y), and E(XY)

But I was wondering if I could decompose the problem quickly using properties of expected values, instead of having to do all the math involved manually calculating the expected value?
If so, which properties should I take a look at in order to achieve this? Thanks!
 
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  • #2


Ok I gave it a crack using basic expansion, lemmy know if I am close :)

E(X2(3Y-1))
E(3X2Y-X2)
3E(X2Y)-E(X2)

Then from there its a fairly simple process of taking the integrals of the various functions. In my examples case the marginal prob dens functions are only defined for certain values so all I got to do is E = 3sum(X2.y(x)) - sum(X2.y(x))and badabing.

Yay or nay? :D
 
  • #3


adamwitt said:

Homework Statement



What is the Expected Value of E(X2(3Y-1))

Homework Equations



Properties of Expected Values.
E(X+c) = E(X) + c
E(X+Y) = E(X) + E(Y)


The Attempt at a Solution



I tried googling for a few of the properties I could use but I'm not sure which ones to use, or maybe I haven't found the required property?
I have been given the data and worked out the marginal probability mass functions so I can work out E(X), E(Y), and E(XY)

But I was wondering if I could decompose the problem quickly using properties of expected values, instead of having to do all the math involved manually calculating the expected value?
If so, which properties should I take a look at in order to achieve this? Thanks!

You need to work out E(X^2*Y). If X and Y are independent, this would be E(X^2)*E(Y), but it they are correlated, there is no simple expression. By the way, you say you have worked out the marginal probability mass functions so can work out E(XY). This statement (the "so can ..." part) is false if X and Y are not independent (or, at least, uncorrelated), because in that case you need the full bivariate probability mass function f(x,y) in order to work out E(XY).

RGV
 

FAQ: How Can You Calculate the Expected Value E(X^2(3Y-1))?

What is the expected value of E(X^2(3Y-1))?

The expected value of E(X^2(3Y-1)) is the average value that we would expect to obtain if we repeatedly calculated X^2(3Y-1) over and over again. This is calculated by taking the sum of all possible outcomes multiplied by their respective probabilities.

How is the expected value of E(X^2(3Y-1)) calculated?

The expected value of E(X^2(3Y-1)) is calculated by first finding the probability distribution of X^2(3Y-1). This can be done by multiplying the probability distribution of X^2 with the probability distribution of 3Y-1. Then, the expected value is calculated by taking the sum of all possible outcomes multiplied by their respective probabilities.

What is the significance of calculating the expected value of E(X^2(3Y-1))?

Calculating the expected value of E(X^2(3Y-1)) allows us to make predictions about the average outcome of X^2(3Y-1) and can help in decision making processes. It is also an important concept in probability theory and is used in various fields such as statistics, finance, and economics.

Can the expected value of E(X^2(3Y-1)) be negative?

Yes, the expected value of E(X^2(3Y-1)) can be negative. This can occur if the outcomes of X^2(3Y-1) are mostly negative and have a higher probability of occurring. However, the expected value is still a useful measure for understanding the average outcome and does not necessarily reflect the exact outcome of any individual calculation.

How can the expected value of E(X^2(3Y-1)) be interpreted?

The expected value of E(X^2(3Y-1)) can be interpreted as the long-term average outcome of X^2(3Y-1) if we were to repeat the calculation multiple times. It is not a guarantee of the exact outcome, but rather a measure of what we can expect to happen on average.

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