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by **zorro** on Sat Jun 26, 2010 6:21 am

Problem : If the probability of X is given by
$f(x) = \begin{cases} x/2, & 0 < x < 1 \\ 1/2, & 1 < x < 2 \\ (3-x)/2, & 2 < x < 3 \\ 0 & elsewhere \end{cases}$

find the expected value of $g(x) = X^2 - 5X + 3$

Solution :

$\int\limits_{0}^{1} . g(x) \frac{x}{2}, dx + \int\limits_{1}^{2} . g(x) \frac{1}{2}, dx + \int\limits_{2}^{3} g(x) \frac{3-x}{2}, dx$

is this process which i am following is it correct or no?

**Sponsor**

by **jk22** on Sat Jun 26, 2010 1:43 pm

Hallo, nice to meet you. >I'm new on this forum.

For me this process of calculation is correct, even if there is a way to see the things that I'm not sure about :

We need <g(X)=Y>=<Y>, and then we need the probability density of Y. I don't know if this gives the same result. (We know that Y runs over [-13/4;3] if x runs over [0;3]).

From this we see there could appear a problem if g were not bijective.

See you soon, with friendly greetings.

by **maggiemagnet** on Sat Jun 26, 2010 1:47 pm

According to this Wikipedia article, yes, it looks like you have set this up correctly. Do you need any help with the integrations?

by **jk22** on Sat Jun 26, 2010 2:47 pm

Hallo, nice to meet you too.

Thanks for the reference. Me I lost myself in : should I calculate like this :

y=g(x), then :

$<g>=<y>=\int yf_y(y)dy=\int_3^1yf_y(y)dy+...\textrm{other sections}=\int_0^1 g(x)f_y(g(x))g'(x)dx+...=\int_0^1g(x)f(x)dx+...$

and hence we find the distribution of Y : f_y(g(x))g'(x)=f(x), or f_y(y)=f(g^-1(y))/g'(g^-1(y)), to get the same as above.

by **zorro** on Sun Jun 27, 2010 7:23 am

THANKS GUYS FOR ALL UR POST ......I JUST NEEDED TO CHECK IF I WAS DOING IT CORRECTLY OR NO ...... :clap:

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