sum of two discrete uniform random variables
Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.
My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$
I'm stuck here and don't know how to proceed. What other cases are needed to be considered?
probability
asked Nov 28 '18 at 4:32
dxdydz
1949
add a comment |
Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.
My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$
I'm stuck here and don't know how to proceed. What other cases are needed to be considered?
probability
asked Nov 28 '18 at 4:32
dxdydz
1949
Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08
add a comment |
Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.
My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$
I'm stuck here and don't know how to proceed. What other cases are needed to be considered?
probability
asked Nov 28 '18 at 4:32
dxdydz
1949
Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.
My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$
I'm stuck here and don't know how to proceed. What other cases are needed to be considered?
probability
probability
asked Nov 28 '18 at 4:32
dxdydz
1949
asked Nov 28 '18 at 4:32
dxdydz
1949
asked Nov 28 '18 at 4:32
dxdydz
1949
asked Nov 28 '18 at 4:32
dxdydz
1949
asked Nov 28 '18 at 4:32
dxdydz
1949
1949
Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08
add a comment |
Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08
Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08
Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08
add a comment |
1 Answer
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You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$
To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$
So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).
In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
add a comment |
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1 Answer
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1 Answer
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You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$
To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$
So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).
In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
add a comment |
You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$
To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$
So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).
In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
add a comment |
You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$
To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$
So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).
In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$
To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$
So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).
In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
edited Nov 28 '18 at 6:01
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
answered Nov 28 '18 at 5:12
Alejandro Nasif Salum
4,399118
4,399118
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
add a comment |
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01
add a comment |
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Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08