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Property of independent random variables

alfred

New member
Apr 30, 2013
10
hello!

I'm trying to understand the following property:
Let X and Y be independent random variables z: = X + Y. Then

where fZ (z) is the probability mass function for a discrete random variable defined as follows:



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

Proof: Using the Law of total probability, which is this:

we obtain

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:

For 2 <= z <= 7 we obtain

And 7 <z <= 12:

Here I just get that:

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!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,721
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:

For 2 <= z <= 7 we obtain

And 7 <z <= 12:

Here I just get that:

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 \(\displaystyle \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36}\) 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

New member
Apr 30, 2013
10
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,721
Okay, so we have got as far as \(\displaystyle \text{Pr}[Z=z] = \frac16 \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\text{Pr}[Y= z-x]\). 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 \(\displaystyle \text{Pr}[Z=z] = \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36} \). 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 \(\displaystyle \frac{z-1}{36}\). 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 \(\displaystyle \frac{13-z}{36}\).
 

alfred

New member
Apr 30, 2013
10
Okay, so we have got as far as \(\displaystyle \text{Pr}[Z=z] = \frac16 \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\text{Pr}[Y= z-x]\). 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 \(\displaystyle \text{Pr}[Z=z] = \sum_{x=\max\{1,z-6\}}^{\min\{6,z-1\}}\frac1{36} \). 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 \(\displaystyle \frac{z-1}{36}\). 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 \(\displaystyle \frac{13-z}{36}\).
Thank you very much! I understand it with your explanation. =D