Integration Problem: Showing Integral Equals 1

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In summary: The second method is the one I was trying to do. I have the book "Schaum's Outline of Theory and Problems of Probability, Random Variables, and Random Processes" by Hwei Hsu and there is a section in there about transforming random variables. The problem I am working on was a problem set in the "Normal Distribution" section. The problem asked to show that \int_0^{\infty} x \frac 1{\sqrt {2\pi}} e^{\frac {-x^2}2} dx = 1by using a transformation of the random variable. The transformation of the random variable is Y=X^2 , where X is a standard random variable. I'm supposed to
  • #1
Mogarrr
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Homework Statement


I'm trying to show that the definite integral:

[itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy [/itex],

equals 1.

Homework Equations



it's already known that [itex] \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy = 1 [/itex], since f(x) is a probability density function.

The Attempt at a Solution


I've tried integration by parts, but that hasn't helped.

[itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy = \frac 1{\sqrt{2 \pi}} (-2 \sqrt {y} e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy [/itex],

but [itex]lim_{t \to \infty} -2 \sqrt {y} e^{\frac {-y}2})|_0^{\infty} = 0 [/itex], so that's not very helpful.

Any ideas to evaluate the integral?
 
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  • #2
Mogarrr said:

Homework Statement


I'm trying to show that the definite integral:

[itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy [/itex]

equals 1.

Homework Equations



it's already known that [itex] \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy = 1 [/itex], since f(x) is a probability density function.

The Attempt at a Solution


I've tried integration by parts, but that hasn't helped.

[itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy = \frac 1{\sqrt{2 \pi}} (-2 \sqrt {y} e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy [/itex],

but [itex]lim_{t \to \infty} -2 \sqrt {y} e^{\frac {-y}2})|_0^{t} = 0 [/itex], so that's not very helpful.
(Fixed a little typo there.)​

Any ideas to evaluate the integral?
If you know that [itex] \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy = 1 \ ,\ [/itex] then it seems to me that you have all but arrived at the result.
 
  • #3
I believe Mogarr knows that it equals 1, though doesn't analytically know how to demonstrate that with standard integration techniques.
 
  • #4
pixatlazaki said:
I believe Mogarr knows that it equals 1, though doesn't analytically know how to demonstrate that with standard integration techniques.

That's right
 
  • #5
Mogarrr said:
That's right
So, which part of this problem are you having difficulty with.

Please be specific.
 
  • #6
SammyS said:
So, which part of this problem are you having difficulty with.

Please be specific.

I'm having a problem finding a method to evaluate:

[itex] \int_0^{\infty} y \frac 1{\sqrt {2\pi}} \cdot y^{\frac {-1}2} e^{\frac {-y}2} dy =\int_0^{\infty} \frac 1{\sqrt {2\pi}} \cdot y^{\frac 12} e^{\frac {-y}2} dy [/itex].

Here's some background. I know that the integral should evaluate to 1. In a previous part of the problem, I saw that for this standard normal function,

[itex] f_X(x) = (\frac 1{\sqrt {2\pi}}) e^{\frac {-x}2} [/itex],

the expected value of the random variable squared, [itex] E X^2 [/itex], is equal to 1. I had to look at the answer to solve this problem. The integral I am trying to solve is the expected value of Y, [itex] E Y [/itex], where [itex] Y = X^2 [/itex]. So this is a transformation of the random variable, but the result should be the same as [itex] E X^2 [/itex] (this was stated in the problem description).

The conundrum I have is, after seeing the answer, I don't believe it to be true. Here is the answer given (hopefully without typos) in equations and italics:

[itex] E Y = \int_0^{\infty} \frac {Y}{\sqrt {2\pi y}}e^{\frac {-y}2} dy = \frac 1{2\pi} (-2y^{\frac 12}e^{\frac {-y}2}|_0^{\infty} + \int_0^{\infty} y^{\frac {-1}2}e^{\frac {-y}2}dy) = \frac 1{\sqrt {2\pi}} \cdot \sqrt {2\pi} = 1[/itex].

This was obtained using integration by parts with [itex]u=2y^{\frac 12}[/itex] and [itex]dv= \frac 12 e^{\frac {-y}2} [/itex], and the fact that [itex] f_Y(y) [/itex] integrates to 1.

I don't think this answer is right though.

Isn't [itex] -2y^{\frac 12} e^{\frac {-y}2} |_0^{\infty} = 0[/itex] ? And if I am right, then I am fresh out of ideas to solve this problem.
 
  • #7
If we make a simple substitution, [itex] u = \sqrt{y} [/itex], then we can see that $$\int \frac{1}{2\sqrt{y}}e^{-y/2}dy = \int e^{-u^2/2}du $$, what does that tell you about the integral itself?
 
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  • #8
I think I'm getting really close, but thus far...

[itex] \int_0^{\infty} \sqrt {y} e^{\frac {-y}2} dy = 2\sqrt {y} (-e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex]

where I did integration by parts with [itex] u = 2\sqrt {y}[/itex] and [itex] dv = \frac 12e^{\frac {-y}2} dy [/itex].

The first part of the sum is 0. So continuing on...

[itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex] can be simplified with a u-substitution. If I let [itex] u = \sqrt {y} [/itex], then [itex] du = \frac 12 y^{\frac {-1}2}dy [/itex], and the limits of integration won't change.

So [itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy = 2 \int_0^{\infty} e^{-y/2} \frac 12 \frac 1{\sqrt {y}} dy = 2\int_0^{\infty} e^{\frac {-(u)^2}2}du [/itex],

but [itex] 2\int_0^{\infty} e^{\frac {-(u)^2}2}du = 2 \cdot \sqrt {2\pi}[/itex], where I need it to be [itex] \sqrt {2\pi} [/itex] so the whole thing can evaluate to 1.

I think a mistake was made. Don't know where though.
 
Last edited:
  • #9
Mogarrr said:
I think I'm getting really close, but thus far...

[itex] \int_0^{\infty} \sqrt {y} e^{\frac {-y}2} dy = 2\sqrt {y} (-e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex]

where I did integration by parts with [itex] u = 2\sqrt {y}[/itex] and [itex] dv = \frac 12e^{\frac {-y}2} dy [/itex].

The first part of the sum is 0. So continuing on...

[itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex] can be simplified with a u-substitution. If I let [itex] u = \sqrt {y} [/itex], then [itex] du = \frac 12 y^{\frac {-1}2}dy [/itex], and the limits of integration won't change.

So [itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy = 2 \int_0^{\infty} e^{-y/2} \frac 12 \frac 1{\sqrt {y}} dy = 2\int_0^{\infty} e^{\frac {-(u)^2}2}du [/itex],

but [itex] 2\int_0^{\infty} e^{\frac {-(u)^2}2}du = 2 \cdot \sqrt {2\pi}[/itex], where I need it to be [itex] \sqrt {2\pi} [/itex] so the whole thing can evaluate to 1.

I think a mistake was made. Don't know where though.

You can do it in two ways:
(1) recognize that ##\sqrt{y} \exp(-y/2) = y^{3/2 - 1} \exp(-y/2)## is proportional to the density function of the Gamma random variable G(3/2,1/2); see, eg., http://en.wikipedia.org/wiki/Gamma_distribution . You can then reduce the integral to a something involving ##\Gamma(3/2) = (1/2) \Gamma(1/2)## and use the known value of ##\Gamma(1/2)## (or work it out yourself).
(2) Use the change of variable ##y = x^2## to get ##c \int_0^{\infty} x^2 \exp(-x^2/2) \, dx## for a constant ##c##; then you need to know that the standard normal random variable has variance = 1, as you have already mentioned. The easiest way to get that is to use integration by parts:
[tex] \int x^2 e^{-x^2/2} \, dx = \int u dv, \\
u = x, \: dv = x e^{-x^2/2} \, dx = d \left(- e^{-x^2/2} \right)[/tex]
 
  • #10
Mogarrr said:
but [itex] 2\int_0^{\infty} e^{\frac {-(u)^2}2}du = 2 \cdot \sqrt {2\pi}[/itex], where I need it to be [itex] \sqrt {2\pi} [/itex] so the whole thing can evaluate to 1.

I found my mistake. I found this from Wikipedia:

[itex] \int_0^{\infty} e^{-ax^{b}}dx = \frac 1{b} a^{\frac {-1}{b}} \Gamma (\frac 1{b})[/itex]

So after doing looking back at the u-substitution, I completed the problem. Thanks.
 

1. What is an integration problem?

An integration problem involves finding the area under a curve between two given points. It is a fundamental concept in calculus and is used to solve a variety of real-world problems.

2. How do you solve an integration problem?

To solve an integration problem, you must use integration techniques such as substitution, integration by parts, or partial fractions. These techniques help to simplify the integral and make it easier to solve.

3. What does it mean to show an integral equals 1?

Showing an integral equals 1 means that the area under the curve between the given points is equal to 1. This can be achieved by integrating a function and setting the resulting integral equal to 1.

4. Why is it important to solve integration problems?

Solving integration problems is important because it allows us to find the area under a curve, which has many real-world applications. It is also a crucial step in understanding more complex mathematical concepts and solving higher-level problems.

5. What are some tips for solving integration problems?

Some tips for solving integration problems include being familiar with various integration techniques, practicing regularly, and breaking the problem down into smaller, more manageable steps. It is also helpful to check your answers using a calculator or graphing tool.

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