MHB Property of independent random variables

AI Thread Summary
The discussion focuses on the property of the sum of independent random variables, specifically how to derive the probability mass function for Z = X + Y, where X and Y are the outcomes of rolling two dice. The Law of Total Probability is utilized to express the probability of Z in terms of the probabilities of X and Y. The key challenge is understanding the summation limits for X, which depend on the fixed value of Z, leading to conditions that ensure valid outcomes for both dice. The final results show that for different ranges of Z, the probabilities can be calculated based on the number of valid combinations of X and Y. The explanation clarifies the initial confusion regarding the summation index and the reasoning behind the min and max conditions.
alfred2
Messages
8
Reaction score
0
hello!

I'm trying to understand the following property:
Let X and Y be independent random variables z: = X + Y. Then
http://imageshack.us/a/img268/9228/71pe.png
where fZ (z) is the probability mass function for a discrete random variable defined as follows:

http://imageshack.us/a/img801/9218/q71.png

and Wx is the set of possible values ​​for the random variable X

Proof: Using the Law of total probability, which is this:
http://imageshack.us/a/img198/8461/qbn.png
we obtain
http://imageshack.us/a/img855/2067/mbd4.png
Question-I don't see how Law of total probability helps us. I even don't understand how we get the first line of the proof-End of Question

Then my book proposes this example: We roll two dices. Let X/Y random variables, which indicate the number of points in the first/second die. We calculate the density of Z: = X + Y:
http://imageshack.us/a/img10/7541/5fx.png
For 2 <= z <= 7 we obtain
http://imageshack.us/a/img17/9047/stvo.png
And 7 <z <= 12:
http://imageshack.us/a/img90/5522/fuz.png
Here I just get that:
http://imageshack.us/a/img593/7442/ok0i.png
But I do not understand why we have this index of summation nor why we put min and max in the next step
Could you help me please? Thank you!
 
Mathematics news on Phys.org
alfred said:
Then my book proposes this example: We roll two dices. Let X/Y random variables, which indicate the number of points in the first/second die. We calculate the density of Z: = X + Y:
http://imageshack.us/a/img10/7541/5fx.png
For 2 <= z <= 7 we obtain
http://imageshack.us/a/img17/9047/stvo.png
And 7 <z <= 12:
http://imageshack.us/a/img90/5522/fuz.png
Here I just get that:
http://imageshack.us/a/img593/7442/ok0i.png
But I do not understand why we have this index of summation nor why we put min and max in the next step
Could you help me please? Thank you!
The number of spots on the first die is $x$, and on the second die is $y$. You want to know when the sum $x+y$ (the total number of spots on the two dice) is equal to $z$.

If the sum of the spots on the two dice is to be $z$, given that $X=x$, then obviously $x$ must be at least $1$ (because that is the smallest possible value for $x$). But also $x$ must be at least $z-6$ (because $y = z-x$, and $y$ cannot be larger than $6$). Thus we must have $x\geqslant \max\{1,z-6\}$. Next, $x$ cannot be bigger than $6$ (obviously), but also $x$ must not be bigger than $z-1$ (because $y = z-x$, and $y$ cannot be less than $1$). Thus we must have $x\leqslant \min\{6,z-1\}$. Provided both those conditions hold, there will then be a probability of $1/6$ that $y$ will be equal to $z-x$. That is where the expression $$\sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36}$$ comes from.

I think that if you follow this example carefully, you will begin to see how the proof in the first part of your post works.
 
alfred said:
Okay, so we have got as far as $$\text{Pr}[Z=z] = \frac16 \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\text{Pr}[Y= z-x]$$. In that formula, $z$ is fixed. Once we are inside the summation, $x$ is also fixed, because at that stage we are dealing with what happens for a particular value of $x$. So $\text{Pr}[Y= z-x]$ is the probability that $y$ takes the fixed value $z-x$. And of course the probability that $y$ takes any given single value (in the range 1,...,6) is 1/6.

That gives the formula $$\text{Pr}[Z=z] = \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36} $$. Thus each term in the sum is equal to 1/36, and to evaluate the sum we have to multiply 1/36 by the number of terms. If $z$ lies between 2 and 7, then $\max\{1,z-6\} = 1$ and $\min\{6,z-1\} = z-1$. So the sum goes from $x=1$ to $x=z-1$. Hence there are $z-1$ terms in the sum, and since each term is equal to 1/36, the sum is $$\frac{z-1}{36}$$. In a similar way, you should be able to work out that if $z$ lies between 7 and 12 then the number of terms in the sum is $13-z$ and so their sum is $$\frac{13-z}{36}$$.
 
Opalg said:
Okay, so we have got as far as $$\text{Pr}[Z=z] = \frac16 \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\text{Pr}[Y= z-x]$$. In that formula, $z$ is fixed. Once we are inside the summation, $x$ is also fixed, because at that stage we are dealing with what happens for a particular value of $x$. So $\text{Pr}[Y= z-x]$ is the probability that $y$ takes the fixed value $z-x$. And of course the probability that $y$ takes any given single value (in the range 1,...,6) is 1/6.

That gives the formula $$\text{Pr}[Z=z] = \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36} $$. Thus each term in the sum is equal to 1/36, and to evaluate the sum we have to multiply 1/36 by the number of terms. If $z$ lies between 2 and 7, then $\max\{1,z-6\} = 1$ and $\min\{6,z-1\} = z-1$. So the sum goes from $x=1$ to $x=z-1$. Hence there are $z-1$ terms in the sum, and since each term is equal to 1/36, the sum is $$\frac{z-1}{36}$$. In a similar way, you should be able to work out that if $z$ lies between 7 and 12 then the number of terms in the sum is $13-z$ and so their sum is $$\frac{13-z}{36}$$.

Thank you very much! I understand it with your explanation. =D
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top