Finding a probability by conditioning
Suppose $X_i$ $i=1,2,3..$ are indepednent random variables with common distribution F(x). Let $N$ be a geometric with parameter $alpha$ and suppose $X_i$ is independent of $N$ for all $i$. Let $M = max(X_1,...,X_N)$. Find $P(X leq x ) $ by conditioning on $N$ and find it in different way (Hint: for second part compute $P(M leq x mid N = 1)$ and $P(M leq x mid N > 1 $). )
Attempt:
By conditioning, we have
$$ P(M leq x) = sum_{i=1}^{infty} P(M leq x mid N=i ) P(N=i) $$
$$ = sum frac{ P(M leq x cap N = i ) cdot P(X=i) }{P(X=i)} = sum P( X_1 leq x, X_2 leq x,...,X_N leq x, N=i)$$
since they are all indepdenent, we have
$$ sum F(x)^N P(N=i) = F^N(x) sum P(N=i) = F^N(x) $$
Now, for the second part, Im not quite sure what they mean. Notice by independence, we have
$$ P(M leq x mid N = 1) = P(M leq x ) $$
$$ P(M leq x mid N > 1) = P(M leq x )$$
am I misunderstanding the hint?
probability
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
add a comment |
Suppose $X_i$ $i=1,2,3..$ are indepednent random variables with common distribution F(x). Let $N$ be a geometric with parameter $alpha$ and suppose $X_i$ is independent of $N$ for all $i$. Let $M = max(X_1,...,X_N)$. Find $P(X leq x ) $ by conditioning on $N$ and find it in different way (Hint: for second part compute $P(M leq x mid N = 1)$ and $P(M leq x mid N > 1 $). )
Attempt:
By conditioning, we have
$$ P(M leq x) = sum_{i=1}^{infty} P(M leq x mid N=i ) P(N=i) $$
$$ = sum frac{ P(M leq x cap N = i ) cdot P(X=i) }{P(X=i)} = sum P( X_1 leq x, X_2 leq x,...,X_N leq x, N=i)$$
since they are all indepdenent, we have
$$ sum F(x)^N P(N=i) = F^N(x) sum P(N=i) = F^N(x) $$
Now, for the second part, Im not quite sure what they mean. Notice by independence, we have
$$ P(M leq x mid N = 1) = P(M leq x ) $$
$$ P(M leq x mid N > 1) = P(M leq x )$$
am I misunderstanding the hint?
probability
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
1
In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40
add a comment |
Suppose $X_i$ $i=1,2,3..$ are indepednent random variables with common distribution F(x). Let $N$ be a geometric with parameter $alpha$ and suppose $X_i$ is independent of $N$ for all $i$. Let $M = max(X_1,...,X_N)$. Find $P(X leq x ) $ by conditioning on $N$ and find it in different way (Hint: for second part compute $P(M leq x mid N = 1)$ and $P(M leq x mid N > 1 $). )
Attempt:
By conditioning, we have
$$ P(M leq x) = sum_{i=1}^{infty} P(M leq x mid N=i ) P(N=i) $$
$$ = sum frac{ P(M leq x cap N = i ) cdot P(X=i) }{P(X=i)} = sum P( X_1 leq x, X_2 leq x,...,X_N leq x, N=i)$$
since they are all indepdenent, we have
$$ sum F(x)^N P(N=i) = F^N(x) sum P(N=i) = F^N(x) $$
Now, for the second part, Im not quite sure what they mean. Notice by independence, we have
$$ P(M leq x mid N = 1) = P(M leq x ) $$
$$ P(M leq x mid N > 1) = P(M leq x )$$
am I misunderstanding the hint?
probability
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
Suppose $X_i$ $i=1,2,3..$ are indepednent random variables with common distribution F(x). Let $N$ be a geometric with parameter $alpha$ and suppose $X_i$ is independent of $N$ for all $i$. Let $M = max(X_1,...,X_N)$. Find $P(X leq x ) $ by conditioning on $N$ and find it in different way (Hint: for second part compute $P(M leq x mid N = 1)$ and $P(M leq x mid N > 1 $). )
Attempt:
By conditioning, we have
$$ P(M leq x) = sum_{i=1}^{infty} P(M leq x mid N=i ) P(N=i) $$
$$ = sum frac{ P(M leq x cap N = i ) cdot P(X=i) }{P(X=i)} = sum P( X_1 leq x, X_2 leq x,...,X_N leq x, N=i)$$
since they are all indepdenent, we have
$$ sum F(x)^N P(N=i) = F^N(x) sum P(N=i) = F^N(x) $$
Now, for the second part, Im not quite sure what they mean. Notice by independence, we have
$$ P(M leq x mid N = 1) = P(M leq x ) $$
$$ P(M leq x mid N > 1) = P(M leq x )$$
am I misunderstanding the hint?
probability
probability
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
asked Nov 30 at 6:06
Jimmy Sabater
1,930219
1,930219
1
In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40
add a comment |
1
In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40
1
1
In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40
add a comment |
1 Answer
1
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begin{align}
P(M le x) &= sum_{n=1}^infty P(M le x|N=n)P(N=n)\
&=sum_{n=1}^infty P(max(X_1, ldots, X_n )le x )(1-alpha)^{n-1}alpha\
&= frac{alpha}{1-alpha}sum_{n=1}^infty [F(x)(1-alpha)]^n \
&= frac{alpha}{1-alpha}frac{F(x)(1-alpha)}{1-F(x)(1-alpha)}\
&= frac{F(x)alpha}{1-F(x)(1-alpha)}
end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=frac{P(N>1+n)}{P(N>1)}=frac{(1-alpha)^{n+1}}{(1-alpha)}=P(N > n)$$
Hence,
begin{align}
P(M le x) &= P(M le x|N=1)P(N=1)+P(M le x|Nge1)P(N>1)\
&= F(x)alpha + (1-alpha) P(M le x|N > 1)\
&= F(x) alpha +(1-alpha)P(X_1 le x)P(max(X_2, ldots, X_N) le x|N > 1)\
&= F(x)alpha + (1-alpha)F(x)P(max(X_1, ldots, X_{N-1} )le x|N >1)\
&= F(x) alpha + (1-alpha) F(x) P(M le x)
end{align}
Hence $$(1-(1-alpha)F(x))P(M le x) = F(x) alpha$$
$$P( M le x) = frac{F(x) alpha}{1-(1-alpha)F(x)}$$
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
|
show 4 more comments
Your Answer
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begin{align}
P(M le x) &= sum_{n=1}^infty P(M le x|N=n)P(N=n)\
&=sum_{n=1}^infty P(max(X_1, ldots, X_n )le x )(1-alpha)^{n-1}alpha\
&= frac{alpha}{1-alpha}sum_{n=1}^infty [F(x)(1-alpha)]^n \
&= frac{alpha}{1-alpha}frac{F(x)(1-alpha)}{1-F(x)(1-alpha)}\
&= frac{F(x)alpha}{1-F(x)(1-alpha)}
end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=frac{P(N>1+n)}{P(N>1)}=frac{(1-alpha)^{n+1}}{(1-alpha)}=P(N > n)$$
Hence,
begin{align}
P(M le x) &= P(M le x|N=1)P(N=1)+P(M le x|Nge1)P(N>1)\
&= F(x)alpha + (1-alpha) P(M le x|N > 1)\
&= F(x) alpha +(1-alpha)P(X_1 le x)P(max(X_2, ldots, X_N) le x|N > 1)\
&= F(x)alpha + (1-alpha)F(x)P(max(X_1, ldots, X_{N-1} )le x|N >1)\
&= F(x) alpha + (1-alpha) F(x) P(M le x)
end{align}
Hence $$(1-(1-alpha)F(x))P(M le x) = F(x) alpha$$
$$P( M le x) = frac{F(x) alpha}{1-(1-alpha)F(x)}$$
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
|
show 4 more comments
begin{align}
P(M le x) &= sum_{n=1}^infty P(M le x|N=n)P(N=n)\
&=sum_{n=1}^infty P(max(X_1, ldots, X_n )le x )(1-alpha)^{n-1}alpha\
&= frac{alpha}{1-alpha}sum_{n=1}^infty [F(x)(1-alpha)]^n \
&= frac{alpha}{1-alpha}frac{F(x)(1-alpha)}{1-F(x)(1-alpha)}\
&= frac{F(x)alpha}{1-F(x)(1-alpha)}
end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=frac{P(N>1+n)}{P(N>1)}=frac{(1-alpha)^{n+1}}{(1-alpha)}=P(N > n)$$
Hence,
begin{align}
P(M le x) &= P(M le x|N=1)P(N=1)+P(M le x|Nge1)P(N>1)\
&= F(x)alpha + (1-alpha) P(M le x|N > 1)\
&= F(x) alpha +(1-alpha)P(X_1 le x)P(max(X_2, ldots, X_N) le x|N > 1)\
&= F(x)alpha + (1-alpha)F(x)P(max(X_1, ldots, X_{N-1} )le x|N >1)\
&= F(x) alpha + (1-alpha) F(x) P(M le x)
end{align}
Hence $$(1-(1-alpha)F(x))P(M le x) = F(x) alpha$$
$$P( M le x) = frac{F(x) alpha}{1-(1-alpha)F(x)}$$
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
|
show 4 more comments
begin{align}
P(M le x) &= sum_{n=1}^infty P(M le x|N=n)P(N=n)\
&=sum_{n=1}^infty P(max(X_1, ldots, X_n )le x )(1-alpha)^{n-1}alpha\
&= frac{alpha}{1-alpha}sum_{n=1}^infty [F(x)(1-alpha)]^n \
&= frac{alpha}{1-alpha}frac{F(x)(1-alpha)}{1-F(x)(1-alpha)}\
&= frac{F(x)alpha}{1-F(x)(1-alpha)}
end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=frac{P(N>1+n)}{P(N>1)}=frac{(1-alpha)^{n+1}}{(1-alpha)}=P(N > n)$$
Hence,
begin{align}
P(M le x) &= P(M le x|N=1)P(N=1)+P(M le x|Nge1)P(N>1)\
&= F(x)alpha + (1-alpha) P(M le x|N > 1)\
&= F(x) alpha +(1-alpha)P(X_1 le x)P(max(X_2, ldots, X_N) le x|N > 1)\
&= F(x)alpha + (1-alpha)F(x)P(max(X_1, ldots, X_{N-1} )le x|N >1)\
&= F(x) alpha + (1-alpha) F(x) P(M le x)
end{align}
Hence $$(1-(1-alpha)F(x))P(M le x) = F(x) alpha$$
$$P( M le x) = frac{F(x) alpha}{1-(1-alpha)F(x)}$$
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
begin{align}
P(M le x) &= sum_{n=1}^infty P(M le x|N=n)P(N=n)\
&=sum_{n=1}^infty P(max(X_1, ldots, X_n )le x )(1-alpha)^{n-1}alpha\
&= frac{alpha}{1-alpha}sum_{n=1}^infty [F(x)(1-alpha)]^n \
&= frac{alpha}{1-alpha}frac{F(x)(1-alpha)}{1-F(x)(1-alpha)}\
&= frac{F(x)alpha}{1-F(x)(1-alpha)}
end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=frac{P(N>1+n)}{P(N>1)}=frac{(1-alpha)^{n+1}}{(1-alpha)}=P(N > n)$$
Hence,
begin{align}
P(M le x) &= P(M le x|N=1)P(N=1)+P(M le x|Nge1)P(N>1)\
&= F(x)alpha + (1-alpha) P(M le x|N > 1)\
&= F(x) alpha +(1-alpha)P(X_1 le x)P(max(X_2, ldots, X_N) le x|N > 1)\
&= F(x)alpha + (1-alpha)F(x)P(max(X_1, ldots, X_{N-1} )le x|N >1)\
&= F(x) alpha + (1-alpha) F(x) P(M le x)
end{align}
Hence $$(1-(1-alpha)F(x))P(M le x) = F(x) alpha$$
$$P( M le x) = frac{F(x) alpha}{1-(1-alpha)F(x)}$$
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
answered Dec 2 at 9:04
Siong Thye Goh
98.9k1464116
98.9k1464116
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
|
show 4 more comments
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
On second line, why $P( M leq x mid N = n ) = P( max (X_1,..,X_n) leq x) $?
– Jimmy Sabater
Dec 3 at 5:28
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
$P(M le x |N=n) = P(max(X_1, ldots, X_N ) le x|N=n) = P(max(X_1, ldots, X_n ) le x|N=n)=P(max(X_1, ldots, X_n ) le x)$
– Siong Thye Goh
Dec 3 at 5:46
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
in the fourth equality on the second derivation of $P(M leq x)$, how does the $X_1$ appear again and how pulled it out of max function?
– Neymar
Dec 4 at 3:47
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
$X_2, ldots, X_N |N > 1$ and $X_1, ldots, X_{N-1}|N>1$ follows the same distribution.
– Siong Thye Goh
Dec 4 at 3:56
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
But Im not given that the $X_i$ are identically distributed.
– Neymar
Dec 4 at 4:00
|
show 4 more comments
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In your first part calculation, note that you are using the dummy variable $i$ in the argument of the pmf $Pr{N = i}$. So when you are simplifying, you should have $F(x)^i$ inside the summand and cannot pull it out.
– BGM
Nov 30 at 6:31
$M$ is not independent of $N$.
– d.k.o.
Dec 2 at 8:40