PDF of a continuous random variable

ThiagoG
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Homework Statement


Let X denote a continuous random variable with probability density function f(x) = kx3/15 for 1≤X≤2. Determine the value of the constant k.

Homework Equations


I'm not sure if this is right but I think ∫kx3/15 dx=1 with the parameters being between 2 and 1,

The Attempt at a Solution


So I did what I i showed in the relevant equations section. I got k = 4. When I plug 2 into the equation for X, my probability is greater than 1 so I know this isn't right. I'm not sure what else to do.
 
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ThiagoG said:
When I plug 2 into the equation for X, my probability is greater than 1 so I know this isn't right.

By this do you mean the PDF f(x) > 1? There is nothing wrong with this as f(x) is the probability density and not a probability in itself.
 
ThiagoG said:

Homework Statement


Let X denote a continuous random variable with probability density function f(x) = kx3/15 for 1≤X≤2. Determine the value of the constant k.

Homework Equations


I'm not sure if this is right but I think ∫kx3/15 dx=1 with the parameters being between 2 and 1,

The Attempt at a Solution


So I did what I i showed in the relevant equations section. I got k = 4. When I plug 2 into the equation for X, my probability is greater than 1 so I know this isn't right. I'm not sure what else to do.

As Orodruin has indicated, f(x) can be greater than 1; the probability that ##X## lies in the interval ##(x,x + \Delta x)## is (for small ##\Delta x > 0##) given by ##P(x,x+\Delta x) = f(x) \cdot \Delta x + o(\Delta x)##. Here, the notation ##o(h)## means terms of higher order in small ##h> 0##; that is ##o(h)/h \to 0## as ##h \to 0##. In other words, for small ##\Delta x > 0## the probability is nearly proportional to ##\Delta x##, with coefficient ##f(x)##. Even if we have, say ##f(x) = 10,## the probability would be ##10 \times 0.0001 = 1/1000 << 1## if ##\Delta x = 0.001##.
 
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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