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[SOLVED] Integral of Probability Distribution Function

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name14

New member
Nov 1, 2012
2
The distance of someone from the center (cable station) of a circle is depicted as r, and the regular radius of that circle(the cable station) is depicted as rc. The Circle represents the entire area the cable station provides cable service for.

Given:
Probability Distribution Function

P(r) = B if r <= rc (the person’s length is shorter than the radius of the cable station so they are in the circle)

P(r) = 0 if r > rc (the person’s length is longer than the radius of the cable station so they are outside the circle, hence distribution factor is 0)

B is a "normalization factor". B is chosen so that probability of finding a r in the circle is 1. For simplicity we have written P as a function of and not both r and θ. We must remember to perform all necessary integration over both r and θ.

We determine B by integrating P(r) from 0 to ∞ (over r) and from 0 to 2pi (over θ).

1 =∫P(r)rdrdθ (I could not figure out how to put zeros under the integral symbols)

Using equations shown above for probability and integration find an expression for B. Also what assumption can I make about the location of someone within that circle.
The function representing the cost of the person who is a cable user:
C(r) =λr where λ is a constant where r is the length from the person to the cable station.

and also C =∫C(r)P(r)rdrdθ (sorry did not know how to put 0 on the bottom of integrals)

I have to derive an expression for C which represent the average cost for each cable station. Any Idea what C would be then? Thanks.

I was told to use ideas of integration and polar coordinates if I like to find the expressions.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
The distance of someone from the center (cable station) of a circle is depicted as r, and the regular radius of that circle(the cable station) is depicted as rc. The Circle represents the entire area the cable station provides cable service for.

Given:
Probability Distribution Function

P(r) = B if r <= rc (the person’s length is shorter than the radius of the cable station so they are in the circle)

P(r) = 0 if r > rc (the person’s length is longer than the radius of the cable station so they are outside the circle, hence distribution factor is 0)

B is a "normalization factor". B is chosen so that probability of finding a r in the circle is 1. For simplicity we have written P as a function of and not both r and θ. We must remember to perform all necessary integration over both r and θ.

We determine B by integrating P(r) from 0 to ∞ (over r) and from 0 to 2pi (over θ).

1 =∫P(r)rdrdθ (I could not figure out how to put zeros under the integral symbols)

Using equations shown above for probability and integration find an expression for B. Also what assumption can I make about the location of someone within that circle.
The function representing the cost of the person who is a cable user:
C(r) =λr where λ is a constant where r is the length from the person to the cable station.

and also C =∫C(r)P(r)rdrdθ (sorry did not know how to put 0 on the bottom of integrals)

I have to derive an expression for C which represent the average cost for each cable station. Any Idea what C would be then? Thanks.

I was told to use ideas of integration and polar coordinates if I like to find the expressions.
Is that the question exactly as asked?

If yes, is there context that you have not mentioned?

CB
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The distance of someone from the center (cable station) of a circle is depicted as r, and the regular radius of that circle(the cable station) is depicted as rc. The Circle represents the entire area the cable station provides cable service for.

Given:
Probability Distribution Function

P(r) = B if r <= rc (the person’s length is shorter than the radius of the cable station so they are in the circle)

P(r) = 0 if r > rc (the person’s length is longer than the radius of the cable station so they are outside the circle, hence distribution factor is 0)

B is a "normalization factor". B is chosen so that probability of finding a r in the circle is 1. For simplicity we have written P as a function of and not both r and θ. We must remember to perform all necessary integration over both r and θ.

We determine B by integrating P(r) from 0 to ∞ (over r) and from 0 to 2pi (over θ).

1 =∫P(r)rdrdθ (I could not figure out how to put zeros under the integral symbols)

Using equations shown above for probability and integration find an expression for B. Also what assumption can I make about the location of someone within that circle.
The function representing the cost of the person who is a cable user:
C(r) =λr where λ is a constant where r is the length from the person to the cable station.

and also C =∫C(r)P(r)rdrdθ (sorry did not know how to put 0 on the bottom of integrals)

I have to derive an expression for C which represent the average cost for each cable station. Any Idea what C would be then? Thanks.

I was told to use ideas of integration and polar coordinates if I like to find the expressions.
Hi name14, :)

For the first integration,

\[1=\int_{0}^{2\pi}\int_{0}^{\infty}P(r)r\,dr\,d \theta\]

Since,

\[P(r) = \begin{cases} B & \mbox{if } r \leq r_c \\ 0 & \mbox{if } r > r_c \end{cases}\]

we have,

\[1=\int_{0}^{2\pi}\int_{0}^{\infty}P(r)r\,dr\,d \theta=\int_{0}^{2\pi}\int_{0}^{r_c}Br\,dr\,d \theta\]

\[\Rightarrow 1=B\int_{0}^{2\pi}\frac{r_{c}^{2}}{2}\,d \theta\]

\[\Rightarrow 1=B\pi r_{c}^{2}\]

\[\therefore B=\frac{1}{\pi r_{c}^{2}}\]

For the second integral,

\begin{eqnarray}

C&=&\int_{0}^{2\pi}\int_{0}^{\infty}C(r)P(r)r\,dr \,d \theta\\

&=&\lambda\int_{0}^{2\pi}\int_{0}^{\infty}P(r)r^2 \,dr\,d \theta\\

&=&B\lambda\int_{0}^{2\pi}\int_{0}^{r_c}r^2 \,dr\,d \theta\\

&=&\frac{\lambda}{\pi r_{c}^{2}}\int_{0}^{2\pi}\int_{0}^{r_c}r^2 \,dr\,d \theta\\

&=&\frac{\lambda}{\pi r_{c}^{2}}\left(\frac{2\pi r_{c}^{3}}{3}\right)\\

&=&\frac{2\lambda r_{c}}{3}\\

\end{eqnarray}

Kind Regards,
Sudharaka.
 
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