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kingwinner
11-21-07, 12:57 AM
Q: Given f(x) = cx + (c^2)(x^2), 0<x<1.
What is c such that the above is a proper probability density function?

Solution:
1
∫ f(x) dx = 1
0
=> 2(c^2) + 3c - 6 =0
=> c= (-3 + sqrt57) / 4 or c= (-3 - sqrt57) / 4
=> Answer: c= (-3 + sqrt57) / 4 (the second one rejected)
======================================

Now, what is the reason of rejecting c= (-3 - sqrt57) / 4 ?

Also, in which step is this extraneous solution produced and why is it produced?

Thanks for explaining!

Reiku
11-21-07, 01:01 AM
I think they might whip your ass first for not putting it in Latex... Lol... :)

BenTheMan
11-21-07, 01:05 AM
indeed...

BenTheMan
11-21-07, 01:06 AM
ummm... did you plug it in and see what happens?

kingwinner
11-21-07, 01:09 AM
I think they might whip your ass first for not putting it in Latex... Lol... :)
Well...I don't know how to use Latex...
I looked at the thread talking about Latex, and it seems really complicated to me...I haven't studied any computer programming stuff...

So sorry, but I think this question is short enough to be readable...

BenTheMan
11-21-07, 01:11 AM
I looked at the thread talking about Latex, and it seems really complicated to me

Well, it's not. You might try to learn how to use it before giving up...

przyk
11-21-07, 01:42 PM
I looked at the thread talking about Latex, and it seems really complicated to me...
To display \LaTeX here at Sciforums, you type in LaTeX code and surround it in "tex" tags, just like you'd use ... to display text in bold. For example:

E = m c^2

is how you'd display one of Reiku's all time favourite equations.

You can start with simple stuff like x_2 and x^2 to display subscripts ( x_2 ) and superscripts ( x^2 ) respectively. You could also try:

\frac{ abc }{ xyz }

to display a fraction like \frac{ abc }{ xyz }, or:

\sqrt{ abc }

for square roots: \sqrt{ abc }.

Anyway, back to your original question, plug c = - \frac{ 3 + \sqrt{57} }{ 4 } (the wrong answer) into your function and see what happens for values of x close to 0.

Absane
11-21-07, 09:56 PM
Q: Given f(x) = cx + (c^2)(x^2), 0<x<1.
What is c such that the above is a proper probability density function?

Well you should first state that X is a continuous random variable. However, it should be implied by the fact that 0 < x < 1.

Anyway:

\int_{-\infty}^{\infty} c*x + c^{2}*x^{2} dx

= \int_{0}^{1} c*x + c^{2}*x^{2} dx

= [c*\frac{x^{2}}{2} + c^{2}*\frac{x^{3}}{3}]_{0}^{1}

= \frac{c}{2} + \frac{c^{2}}{3}

Since this integal must be equal to 1 (definition of a probability density function), just solve:
\frac{c}{2} + \frac{c^{2}}{3} - 1 = 0

Which you get two answers. Are both correct? Is only one correct? Are neither correct!?

Well, remember that a second condition for a probability density function is that f(x) \geq 0 for all x.

c*x + c^{2}*x^{2} \geq 0

x \geq \frac{-1}{c} or x \geq 0 or c \geq 0

Well the second condition is obviously true! For one of the solutions, c = -\frac{3+\sqrt{54}}{4}, condition 1 fails for some x (for example, x = 0.25 fails) and condition 3 fails completely.

So looking at the second solution, it passes all above.

kingwinner
11-21-07, 10:08 PM
You can start with simple stuff like x_2 and x^2 to display subscripts ( x_2 ) and superscripts ( x^2 ) respectively. You could also try:

x_100 and x^100

kingwinner
11-21-07, 10:09 PM
Well, remember that a second condition for a probability density function is that f(x) \geq 0 for all x.

c*x + c^{2}*x^{2} \geq 0

x \geq \frac{-1}{c} or x \geq 0 or c \geq 0

Well the second condition is obviously true! For one of the solutions, c = -\frac{3+\sqrt{54}}{4}, condition 1 fails for some x (for example, x = 0.25 fails) and condition 3 fails completely.

So looking at the second solution, it passes all above.
Thanks a lot!
But how did you get the condition x \geq \frac{-1}{c} or x \geq 0 or c \geq 0?

Absane
11-21-07, 10:14 PM
Thanks a lot!
But how did you get the condition x \geq \frac{-1}{c} or x \geq 0 or c \geq 0?

Do you remember how to solve a quadratic as an inequality?

All I did was factor it and notice that the factors must all multiply out to be positive.

przyk
11-22-07, 03:54 PM
x_100 and x^100
For multiple digits or complicated expressions, enclose them in curly braces {}:

x_{100} will display x_{100}, and

x^{ 1 + \sqrt{ 2 } } will display x^{ 1 + \sqrt{ 2 } }.