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| − | = [[:Category:Problem_solving|Practice Problem]]: normalizing the probability mass function of a | + | = [[:Category:Problem_solving|Practice Problem]]: normalizing the probability mass function of a continuous random variable= |
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A random variable X has the following probability density function: | A random variable X has the following probability density function: | ||
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===Answer 1=== | ===Answer 1=== | ||
| − | + | By the properties of pdfs we know that: | |
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| + | <math>\int\limits_{-\infty}^{+\infty} f_X(x) dx = 1.</math> | ||
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| + | <math>Let, I = \int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx.</math> | ||
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| + | By introduction a new term <math> e^{\frac{-y^2}{2}} </math> we know that: | ||
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| + | <math>I^2 = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{-\frac{1}{2}\left(x^2 + y^2\right)}dxdy.</math> | ||
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| + | Now, we need to change to polar coordinates. We know that <math> r^2 = x^2 + y^2, </math> and that the Jacobian for Polar coordinates is <math> r. </math> By performing the substitution we have: | ||
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| + | <math> I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr. </math> | ||
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| + | Now, we just need to evaluate the integral: | ||
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| + | <math> I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr = 2\pi\int\limits_{0}^{+\infty}re^{-\frac{1}{2}r^2}dr = 2\pi\left(-e^{-\frac{1}{2}r^2}\right)\bigg|_{0}^{+\infty} = 2\pi\left(0 + 1\right) = 2\pi.</math> | ||
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| + | Now, we solve for <math>I</math>: | ||
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| + | <math>I = \sqrt{2\pi}.</math> | ||
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| + | Returning to the original problem, we actually want to solve: | ||
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| + | <math>1 = \int\limits_{-\infty}^{+\infty} f_X(x) dx = \int\limits_{-\infty}^{+\infty}C e^{\frac{-x^2}{2}}dx = C\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx = C \times I = C \times \sqrt{2\pi}. </math> | ||
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| + | Now, solving for the constant <math> C: </math> | ||
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| + | <math> C = \frac{1}{\sqrt{2\pi}}. </math> | ||
| + | *<span style="color:green">Instructor's comments: Good job! You have a good approach, and all your steps are clearly explained. Now I want to challenge the other students: can you write a shorter but equivalent solution? In order words, can you "compactify" the justification? -pm </span> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | + | <math> f_X (x)</math> is in the form of a PDF of a 1-D gaussian random variable. The general form for the gaussian random variable's PDF is <math>\frac{1}{\sqrt{2\pi}s}e^{-\frac{(x-m)^2}{2s^2}}</math>; therefore, it can be seen from the problem statement that m is zero and <math>s^2</math> is one. The value of C is then <math>\frac{1}{\sqrt{2\pi}s}</math>, which is <math>\frac{1}{\sqrt{2\pi}}</math>, since s is either positive or negative one (and only positive one will yield a logical answer). | |
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
Latest revision as of 08:11, 25 February 2013
Contents
Practice Problem: normalizing the probability mass function of a continuous random variable
A random variable X has the following probability density function:
$ f_X (x) = C e^{\frac{-x^2}{2}} , $
where C is a constant. Find the value of the constant C.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
By the properties of pdfs we know that:
$ \int\limits_{-\infty}^{+\infty} f_X(x) dx = 1. $
$ Let, I = \int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx. $
By introduction a new term $ e^{\frac{-y^2}{2}} $ we know that:
$ I^2 = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{-\frac{1}{2}\left(x^2 + y^2\right)}dxdy. $
Now, we need to change to polar coordinates. We know that $ r^2 = x^2 + y^2, $ and that the Jacobian for Polar coordinates is $ r. $ By performing the substitution we have:
$ I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr. $
Now, we just need to evaluate the integral:
$ I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr = 2\pi\int\limits_{0}^{+\infty}re^{-\frac{1}{2}r^2}dr = 2\pi\left(-e^{-\frac{1}{2}r^2}\right)\bigg|_{0}^{+\infty} = 2\pi\left(0 + 1\right) = 2\pi. $
Now, we solve for $ I $:
$ I = \sqrt{2\pi}. $
Returning to the original problem, we actually want to solve:
$ 1 = \int\limits_{-\infty}^{+\infty} f_X(x) dx = \int\limits_{-\infty}^{+\infty}C e^{\frac{-x^2}{2}}dx = C\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx = C \times I = C \times \sqrt{2\pi}. $
Now, solving for the constant $ C: $
$ C = \frac{1}{\sqrt{2\pi}}. $
- Instructor's comments: Good job! You have a good approach, and all your steps are clearly explained. Now I want to challenge the other students: can you write a shorter but equivalent solution? In order words, can you "compactify" the justification? -pm
Answer 2
$ f_X (x) $ is in the form of a PDF of a 1-D gaussian random variable. The general form for the gaussian random variable's PDF is $ \frac{1}{\sqrt{2\pi}s}e^{-\frac{(x-m)^2}{2s^2}} $; therefore, it can be seen from the problem statement that m is zero and $ s^2 $ is one. The value of C is then $ \frac{1}{\sqrt{2\pi}s} $, which is $ \frac{1}{\sqrt{2\pi}} $, since s is either positive or negative one (and only positive one will yield a logical answer).
Answer 3
Write it here.
