Conditional distribution for random variable on interval

Srumix
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Homework Statement



Find the conditional distribution function and density for the random variable X defined on R given that X is in some interval I = (a,b) where P(X in I) > 0. Assume that the density and distribution for the random variable X is known

Homework Equations



fX|X\inI = P(X\leqx | X\in I) = fX,X\inI(x,x)/fX\inI(x)

The Attempt at a Solution



I'm sorry, but my latex skills are very poor so I will try to describe in words what my problem is.
The problem I'm having is that I know how to calculate the probability of P(X in I) since we just take the integral of the density function over the interval I in question. However, what do I do with the "joint" distribution fX,X\inI(x,x) that I need for the definition of conditional distribution? That is what I can't figure out.

Thanks in advance!
 
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Srumix said:

Homework Statement



Find the conditional distribution function and density for the random variable X defined on R given that X is in some interval I = (a,b) where P(X in I) > 0. Assume that the density and distribution for the random variable X is known

Homework Equations



fX|X\inI = P(X\leqx | X\in I) = fX,X\inI(x,x)/fX\inI(x)

The Attempt at a Solution



I'm sorry, but my latex skills are very poor so I will try to describe in words what my problem is.
The problem I'm having is that I know how to calculate the probability of P(X in I) since we just take the integral of the density function over the interval I in question. However, what do I do with the "joint" distribution fX,X\inI(x,x) that I need for the definition of conditional distribution? That is what I can't figure out.

Thanks in advance!

The conditional density is the coefficient of ##\Delta x## in the first-order (small-##\Delta x##) expansion of
\text{P} \{ x &lt; X &lt; x + \Delta x | a \leq X \leq b \}<br /> = \frac{\text{P} \{ x &lt; X &lt; x+ \Delta x \: \&amp; \: a \leq X \leq b \}}{\text{P} \{ a \leq X \leq b \}}
For ##x \in (a,b)##, can you figure out what is the numerator, in terms of the probability density function ##f(.)##? Can you figure out the denominator?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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