Average distance of the dart (Dartboard)

#1
Hello

I have been struggling with this questions for some time now and I have no idea how to start it. Can you guys give me some tips?

"A large circular dartboard is set up with a "bullseye" at the center of the circle, which is at the coordinate (0,0). A dart is thrown at the center but lands at (X,Y), where X and Y are two different Gaussian random variables. What is the average distance of the dart from the bullseye?"

By average distance I assume when you cast about 1000 times. I know a gaussian random variable can have values from the standard normal distribution. But how do I calculate the average distance?

I am using matlab to solve it. Or im trying atleast.
 

BGM

TS Contributor
#2
First of all you need to define the distance measure here.

In 2-D plane a very common one is the Euclidean distance, which will be

\( \sqrt{X^2 + Y^2} \)

and thus you will need to calculate the expectation

\( E\left[\sqrt{X^2 + Y^2}\right] \)

If \( X, Y \) are two independent standard normal random variables, then we have the nice result

https://en.wikipedia.org/wiki/Normal_distribution#Combination_of_two_independent_random_variables

that the above distance will follow a Rayleigh distribution.
 
#3
First of all you need to define the distance measure here.

In 2-D plane a very common one is the Euclidean distance, which will be

\( \sqrt{X^2 + Y^2} \)

and thus you will need to calculate the expectation

\( E\left[\sqrt{X^2 + Y^2}\right] \)

If \( X, Y \) are two independent standard normal random variables, then we have the nice result

https://en.wikipedia.org/wiki/Normal_distribution#Combination_of_two_independent_random_variables

that the above distance will follow a Rayleigh distribution.
Ye I managed to figure out that it was the "Euclidean distance". I looked at the wikipedia and I read about it. But am I missing something, or how can I calculate a answer from that?

I know that the answer is about 1,2381 (N = 1000).
 

BGM

TS Contributor
#4
You should be already understand the first half of my post, but I am not sure if you missed the second half.

Basically if you are not allowed to use the result that the distance follows a \( \text{Rayleigh}(1) \) distribution, then you need to derive the pdf (actually it is related to \( \chi^2(2) \) distribution, so it is nice) and then do an integration to obtain the expectation.

https://en.wikipedia.org/wiki/Rayleigh_distribution

The expectation is

\( 1 \times \sqrt{\frac {\pi} {2}} \approx 1.25331413732 \)

which is close to the answer you got.