alias said:Homework Statement
Show that E[Y^4] = 3, where Y~N(0,1)
Homework Equations
E[(Y-mu)^4]/[E(Y-mu)^2]^2 = 3
E(Y^4) = 1/sprt(2pi)*intregral (y^4)*e^(-y^2/2)
The Attempt at a Solution
I have expanded and simplified the first equation above and cannot get it to equal 3. I think it's possible to solve the second using integration by parts but I can't find what to use for
[u, du, dv, v] in order to integrate by parts. Any help would be appreciated. Thanks.
alias said:I tried integration by parts but the integral I end up with makes no sense to me.
Integration is a mathematical operation used to calculate the area under a curve. In the context of a probability density function (pdf), integration is used to find the probability of a random variable falling within a certain range of values.
The expected value, also known as the mean, is calculated by integrating the product of a random variable and its corresponding probability from negative infinity to infinity. In other words, it is the weighted average of all possible outcomes of a random variable.
No, integration of a pdf is used to find the probability of a range of values, not a specific outcome. This is because the probability of a specific outcome in a continuous distribution is equal to zero.
The shape of a pdf can affect the integration process by changing the limits of integration and the function being integrated. The limits of integration are determined by the range of values for which the pdf is non-zero, and the function being integrated is the pdf itself.
Yes, there can be limitations to using integration for finding the expected value of a pdf. In some cases, the integral may not converge, meaning that the expected value cannot be calculated. Additionally, integration may not be feasible for complex or high-dimensional pdfs, in which case numerical methods may be used instead.