Question

had to google it but my thoughts are....

Answer 1

http://assets.openstudy.com/updates/attachments/551ed23fe4b0e5665d5f6d80-amistre64-1428084248370-untitled.jpg

Answer 2

X is the age of a car in a crash

E(X) is the expected (age of a car in a crash), had trouble reading that part of it

find x in X that balances the volume, in other words, define f in terms of x alone, integrate out the y parts

\[h(x)=\int_{0}^{1}f(x,y)~dy\]

Answer 3

the mean of X is thereofre just the integration of xh(x)

Answer 4

\[E(X)=\int_{2}^{10}x\left(\int_{0}^{1}\frac1{64}(10-xy^2)~dy\right)~dx\]

Answer 5

it is :) and they might be lol

Answer 6

it takes me longer to read/comprehend the question then anything else .... that and adding seems to thru me for a loop at times.

Answer 7

I think I understand for the most part what's going on, like how you're getting the expectation value makes sense to me, but why are we integrating over y at all? The way I was thinking of it I just would evaluate \[\Large \int\limits_2^{10}xf(x,y)dx\] I guess evaluating over y just gives me... 100% or something?

Answer 8

i wasnt real sure on how to approach it, so i found the solution

the determine f_x (x,y) first, which is the dy integration

Answer 9

the expected value of x is the point at which the volume would balance on the x=k axis

Answer 10

normally, w would find mean as xf(x) but we have a f(x,y) and i spose its some sort of logical progression to just get f(x,y) into a function of x

Answer 11

Yeah, I mean since they're independent variables it doesn't really seem like there should be much issue if there are y's there or not. By the same reasoning I think the order of integration doesn't matter the only problem I am kinda having is I feel like the integral \[\Large \int\limits_0^1\int\limits_2^{10} f(x,y)dxdy\] should equal 1. But it's a made up problem so maybe I'm looking into it too much.

Answer 12

the density function f(x,y) does equal out to 1

Answer 13

Oh so I just messed up my math haha ok

Answer 14

it happens at times ;)

Answer 15

How to power rule? XD

Answer 16

if we had a density function: h(x)

E(X) = integration of x h(x)

Answer 17

so integration of f(x,y) dy gives us a function of x related to the issue

Answer 18

Well we already have a function of x, it just happens to be a function of y as well. We can calculate the expectation of x and not go over all y if we don't want to, we can choose to only look at it over a portion of the total owners, such as those who only insured their cars for up to half a year if we choose to go from 0 to .5 instead I think.

Like if the question said Calculate the expected age of an insured automobile involved in an accident with owners who insured their cars for less than half a year.

I don't know, I'm really just saying this so you can correct me if I'm wrong cause I'm curious and want to learn.

Answer 19

f(x,y) is not necessarily a function of x, and just subsequently a function of y tossed in.



imagine the area of xy divided into strips so that y is held constant

the average of each strip is therfore (x1, x2, x3, x4, ..., xn)

the average of the xs would be the average balancing point in x