PDF of a continuous random variable

Click For Summary
SUMMARY

The discussion focuses on determining the constant k for the probability density function (PDF) f(x) = kx³/15, defined for the continuous random variable X in the interval [1, 2]. The correct approach involves integrating the PDF over the specified interval and setting the integral equal to 1. The integration yields k = 4, which is validated by the understanding that f(x) can exceed 1, as it represents a density rather than a probability. The key takeaway is that the probability of X lying in a small interval is proportional to f(x) multiplied by the width of that interval.

PREREQUISITES
  • Understanding of continuous random variables
  • Knowledge of probability density functions (PDFs)
  • Familiarity with integration techniques
  • Concept of limits and small intervals in probability
NEXT STEPS
  • Study the properties of probability density functions in detail
  • Learn about integration of functions over specified intervals
  • Explore the concept of cumulative distribution functions (CDFs)
  • Investigate applications of continuous random variables in real-world scenarios
USEFUL FOR

Students studying statistics, mathematicians focusing on probability theory, and anyone interested in understanding continuous random variables and their properties.

ThiagoG
Messages
15
Reaction score
0

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.
 
Physics news on Phys.org
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##.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
7
Views
2K
Replies
8
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
9
Views
4K