Probability Density Function

[kingwinner]

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!

 

[HallsofIvy]

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!

$\int_0^1 cx+ c^2x^2 dx= c/2+ c^2/3$. Setting that equal to 1, c/2+ c²/3= 1, and multiplying through by 6, 3c+ 2c²= 6 or 2c²+ 3c- 6= 0. That has, as you say, roots of
$$c= \frac{-3\pm\sqrt{57}}{2}$$

Because the cumulative probabilty must be increasing, if c were negative we could have a section on which the probability from a to b (b> a) might be negative and that is impossible. As to "where the extraneous solution was produced", it was really produced when the "c²" was put into the problem. That introduced the possiblity of a negative value of c.

 

[e(ho0n3]

I don't see why the second one is rejected. Does it say it's rejected or is it just not in the answer?

 

[e(ho0n3]

Ah, OK. Nevermind. Makes sense.

 

[e(ho0n3]

Graphically speaking, if you draw f(x) and notice a section of it below the x-axis, then there's chance of getting a negative probability in some interval [a, b] which is a no no. That means that f must satisfy that additional property that f(x) >= 0 for all values of x. Right?

 

[kingwinner]

Because the cumulative probabilty must be increasing, if c were negative we could have a section on which the probability from a to b (b> a) might be negative and that is impossible.

OK, but how do you know for sure that for the second "c", there IS a section on which the density function is negative? (without a graphing calculator since I am not allowed to use it...)

As to "where the extraneous solution was produced", it was really produced when the "c²" was put into the problem. That introduced the possiblity of a negative value of c.

But the original density function has a c^2 too, I did NOT "square both sides", so there shouldn't be any extraneous solution, right?


Thanks a lot for your help and explanation!

 

[HallsofIvy]

If c= (-3-[itex]\sqrt{57})/4, which is approximately -2.7, then your probability density function is f(x) = cx + (c^2)(x^2) or f(x)= 6.9x²- 2.7x. That's a parabola and has zeros at x= 0 and x= 2.7/6.9= .391! If x= .25, f(.25)= 6.9(.0625)- 2.7(.25)= -.632> 0 which means that f(x)< 0 for all x between 0 and .391. If you were to use that value for c, you would find that the "probability" that x las between, say 0 and .25 was negative!