Prove Pr(t<X<t+dt)=f(t)dt - Get Help Here!

  • Thread starter Thread starter rukawakaede
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving the relationship Pr(t PREREQUISITES

  • Understanding of probability density functions (PDFs)
  • Familiarity with integral calculus and limits
  • Knowledge of the mean value theorem
  • Basic concepts of Stieltjes integration
NEXT STEPS
  • Study the properties of probability density functions (PDFs)
  • Learn about the mean value theorem and its applications in calculus
  • Explore Stieltjes integration and its relevance to probability theory
  • Review the concept of continuity in functions and its implications for integration
USEFUL FOR

Students of mathematics, statisticians, and anyone interested in understanding the foundations of probability theory and integration techniques.

rukawakaede
Messages
58
Reaction score
0
Hi

Could anyone show to me that:

If dt is an infinitely small number, the probability that X is included within the interval (t, t + dt) is equal to f(t) dt ,i.e.
Pr(t&lt;X&lt;t+dt) = f(t)dt
where f is the probability density function.

this sentence is from wikipedia but I could not prove this to myself.

Thanks if anyone can help.
 
Physics news on Phys.org
What are the limits of your integral? You can't really make sense of f(t)dt without some kind of constraint on the upper and lower limits.
 
chiro said:
What are the limits of your integral? You can't really make sense of f(t)dt without some kind of constraint on the upper and lower limits.

here dt is an infinitely small number. and i don't think the RHS is an integral.

maybe you can refer to http://en.wikipedia.org/wiki/Probability_density_function#Further_details

the last line of the Further details
 
Last edited:
The probability density function, f, is defined by the fact that
P(a&lt; x&lt; b)= \int_a^b f(x)dx
or, equivalently,
P(a&lt; x&lt; a+h)= \int_a^{a+h} f(x)dx

What Wikipedia gives is a "differential" form of that.
 
HallsofIvy said:
The probability density function, f, is defined by the fact that
P(a&lt; x&lt; b)= \int_a^b f(x)dx
or, equivalently,
P(a&lt; x&lt; a+h)= \int_a^{a+h} f(x)dx

What Wikipedia gives is a "differential" form of that.

Thanks HallsofIvy for your reply.

I was trying to work out why
\int_a^{a+h} f(x)dx= f(a)h
when h infinitely small.

Could you tell me more about why this is true? or could you explain the differential form? I am still not fully understand yet.
 
rukawakaede said:
Thanks HallsofIvy for your reply.

I was trying to work out why
\int_a^{a+h} f(x)dx= f(a)h
when h infinitely small.
First off, you almost surely do not mean much of what you said literally. You should spend some time thinking about what you actually meant
Anyways, if by "=" you meant "is approximately" and by "infinitely small" you meant "sufficiently small", it's true because of the mean value theorem and the definition of continuous function.

(P.S. if f is not assumed to be continuous, then the equation above is very false)
 
rukawakaede said:
Hi

Could anyone show to me that:

If dt is an infinitely small number, the probability that X is included within the interval (t, t + dt) is equal to f(t) dt ,i.e.
Pr(t&lt;X&lt;t+dt) = f(t)dt
where f is the probability density function.

this sentence is from wikipedia but I could not prove this to myself.

Thanks if anyone can help.

It's just a (dodgy) method for turning sums into integrals. In the long run it's worthwhile to learn Stieltjes integration so you can write probabilities and expectations as
E[g(X)] = \int_R g(x)dF(x)
which is valid whether or not the distribution F(x) has a density.
 

Similar threads

Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K