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corneel
08-27-2010, 09:38 PM
I jus dont seem to understand this concept "conditional e.v."

I do know the definitions: joint density f_X,Y , conditional density of X given Y=y, etc. (from textbook)

Also we have by definition

E( X|Y=y) = int_{R} x f_{X|Y} (x|y) dx ,

with f_{X|Y} (x|y) = f_X,Y (x,y) / f_Y (y) the conditional density of X given Y=y. This is no problem. (calculatting problems for me like E ( X | Y = 3) is not problemfor me..)

But, I have a lots of problems with for example the following :

Let X be a continuous r.v. ; calculatte E( X | a < X< b) .

how is this exactly defined... how to calculatte this, and why ?

I have two option for this kind :

first: E(X| a<X<b) = int_{R} E(X | a<X<b,X=t) f_X(t) dt = int_{a to b} t f_X(t) dt

is this wrong or not ? I do condition there i think.


second : E(X| a<X<b) = ( int_{a tob} t f_X(t) dt ) / P( a < X < b)


could please please someone give to me clear explanation out of definiton how to exactly calculatte this things rigorsly? thank a lot people !

BGM
08-28-2010, 09:23 AM
The second one is correct.

This conditional distribution is a truncated distribution.
You need to get the corresponding conditional p.d.f. by dividing the
probability to normalize it. i.e.

f_{X|a<X<b}(x) = \frac {f_X(x)} {\Pr\{a<X<b\}} , a < X < b

such that \int_a^b f_{X|a<X<b}(x) dx = 1

You can see this by
F_{X|a<X<b}(x) = \Pr\{X < x|a < X < b\}
= \frac {\Pr\{X < x, a < X < b\}} {\Pr\{a<X<b\}}
= \frac {\Pr\{a < X < x\}} {\Pr\{a<X<b\}}, a < x < b
and differentiate it to get the desired result.

Also, if you remember the formula
E[X] = E[X|X \in (a,b)] \Pr\{X \in (a, b)\}
+ E[X|X \notin (a,b)] \Pr\{X \notin (a,b) \}

and compare with
\int_{-\infty}^{+\infty} xf_X(x)dx = \int_a^b xf_X(x)dx +
\left( \int_{-\infty}^a xf_X(x)dx + \int_b^{+\infty} xf_X(x)dx\right) ,

it will help you to remember.

corneel
08-28-2010, 11:39 AM
The second one is correct.

This conditional distribution is a truncated distribution.
You need to get the corresponding conditional p.d.f. by dividing the
probability to normalize it. i.e.

f_{X|a<X<b}(x) = \frac {f_X(x)} {\Pr\{a<X<b\}} , a < X < b

such that \int_a^b f_{X|a<X<b}(x) dx = 1

You can see this by
F_{X|a<X<b}(x) = \Pr\{X < x|a < X < b\}
= \frac {\Pr\{X < x, a < X < b\}} {\Pr\{a<X<b\}}
= \frac {\Pr\{a < X < x\}} {\Pr\{a<X<b\}}, a < x < b
and differentiate it to get the desired result.

Also, if you remember the formula
E[X] = E[X|X \in (a,b)] \Pr\{X \in (a, b)\}
+ E[X|X \notin (a,b)] \Pr\{X \notin (a,b) \}

and compare with
\int_{-\infty}^{+\infty} xf_X(x)dx = \int_a^b xf_X(x)dx +
\left( \int_{-\infty}^a xf_X(x)dx + \int_b^{+\infty} xf_X(x)dx\right) ,

it will help you to remember.

Thank you a lot!! Now it is already more clear for me.

But I have one more thing I don't understand : I made in my first post the calculattion

E(X| a<X<b) = int_{R} E(X | a<X<b,X=t) f_X(t) dt = int_{a to b} t f_X(t) dt

and I do conditioning on there. But result is clearly wrong I know now because what you said. What is exactly wrong with this calculattion then ? i did do wrong conditioning !?

Thanks to you

BGM
08-28-2010, 12:26 PM
It is about the "information" you already conditional on. For example,
E[X] = \int_{-\infty}^{+\infty} E[X|X = x]f_X(x)dx
= \int_{-\infty}^{+\infty} xf_X(x)dx

But
E[X|a<X<b] =
\int_{-\infty}^{+\infty} E[X|a<X<b, X = x]f_{X|a<X<b}(x)dx
= \int_{-\infty}^{+\infty} xf_{X|a<X<b}(x)dx