Chapter 7, Q7.1

[vidhya36]

1. How the limits for fs(s) are decided as 0 to s? [for s between 0 and 10]
2. Why are we considering 10<s<20 and how the limits are decided as s-10 to 10?
3. And, how do we know it is a triangular shaped distribution? We have probability of 1/4 for N=1 which is uniform in 0 to 10. So, shouldn't it be a horizontal straight line till 10, instead of a line sloping up? How the graph is plotted?

 

[suraj]

1. How the limits for fs(s) are decided as 0 to s? [for s between 0 and 10]
2. Why are we considering 10<s<20 and how the limits are decided as s-10 to 10?

First note that we're varying "y" in the integral.

The domain of both X and Y is (0,10) and because S = X + Y , the domain of S is
(0, 20)

Now for 0 < s < 10

the smallest value which y can take is 0 and largest is s.
y can't be more than s, because in that case x has to be negative to satisfy the equality s = x + y

and we know that X does not take negative values.

Similarly for 10 < s < 20

the smallest value which y can take is "s - 10" because if y takes value less than this, then x has to be greater than 10 (which again violates the domain of X) to satisfy the equality s = x + y and the max. value which y can take in this case is 10.

So basically we want to cover all possible outcomes but make sure that X and Y don't take values outside their domain as we're integrating the Joint Density function fxy.

3. And, how do we know it is a triangular shaped distribution?

Plot the PDF. It is triangular shaped symmetric around 10 on the interval (0, 20). Furthermore this is a classic example of "Irwin hall distribution"

https://en.wikipedia.org/wiki/Irwin–Hall_distribution

We have probability of 1/4 for N=1 which is uniform in 0 to 10. So, shouldn't it be a horizontal straight line till 10, instead of a line sloping up? How the graph is plotted?

The continuous part of this distribution is a mixture of two distributions.
First is U(0, 10) and second is a symmetric triangular distribution on interval (0, 20).
So the common part here is the interval (0, 10) which is a combination of a triangle and a rectangle or you can say a triangle sitting on top of a rectangle!

 

[vidhya36]

Thank you very much for your detailed explanation, Sir.
Now, I can see this differently and clearly than before.
One thing is not clear for me.
How did we get plots at 1/20 and 1/40 at the area of convergence i.e., at 10?
The working would be more helpful for me.
Is there any thumb rule to plot area of convergence... Say, in this case, in (0,10) both Uniform and Triangular converges?
(Not sure whether 'converges' is a right word to use.)

 

[suraj]

Sorry for the late reply

(Not sure whether 'converges' is a right word to use.)

Even I'm not sure whether domain was the right word to use. I think the right word is "Range" coz these are RVs not mathematical functions.

Coming back to your doubt,

1/4 times the RV is U(0, 10)

and 1/4 times the RV is a triangular distribution on interval (0, 20)

Now you've to scale down the PDFs of these two RVs by multiplying them with 1/4 and then plot the graph. You can see that the total area under the continuous part is 1/2 and the rest of the area/mass (which is 1/2) is taken by 0 alone.

 

[vidhya36]

Thanks a lot.