Find $P(min{ngt0:X_1+X_2+cdots+X_ngt x}>2)$ for $(X_n)$ i.i.d. uniform on $(0,1)$
Let $X_i$ be i.i.d random variables that follow uniform distributions on $[0,1]$. Let $t(x)=min{ngt0:X_1+X_2+...+X_ngt x}$. Find $P(t(x)gt2)$.
Here's my answer but I'm not sure if I interpreted this question correctly.
$$P(t(x)gt2)=1-P(t(x)le2)$$
$$=1-P(n=1 cup n=2)$$
$$=1-P(X_1gt x)-P(X_1+X_2gt xcap X_1lt x)$$
$$=1-1-0=0. if xin(-infty,0)$$
$$=1-int_{x}^{1}dx_1-intint_{x_1+x_2gt x,x_1lt x}dx_1dx_2. if xin[0,1)$$
$$=1-0-intint_{x_1+x_2gt x}dx_1dx_2. ifxin[1,2) $$
$$=1-0-0=1 if xin[2,infty)$$
I did the computation and the answer looks quite "correct".
My answer is
$0$ if $xin(-infty,0)$.
$x^2over2$ if $xin[0,1)$.
$-x^2+4x-2over2$ if $xin[1,2)$.
$1$ if $xin[2,infty)$
probability uniform-distribution
edited yesterday
Did
246k23221455
asked 2 days ago
Yibei He
1128
add a comment |
Let $X_i$ be i.i.d random variables that follow uniform distributions on $[0,1]$. Let $t(x)=min{ngt0:X_1+X_2+...+X_ngt x}$. Find $P(t(x)gt2)$.
Here's my answer but I'm not sure if I interpreted this question correctly.
$$P(t(x)gt2)=1-P(t(x)le2)$$
$$=1-P(n=1 cup n=2)$$
$$=1-P(X_1gt x)-P(X_1+X_2gt xcap X_1lt x)$$
$$=1-1-0=0. if xin(-infty,0)$$
$$=1-int_{x}^{1}dx_1-intint_{x_1+x_2gt x,x_1lt x}dx_1dx_2. if xin[0,1)$$
$$=1-0-intint_{x_1+x_2gt x}dx_1dx_2. ifxin[1,2) $$
$$=1-0-0=1 if xin[2,infty)$$
I did the computation and the answer looks quite "correct".
My answer is
$0$ if $xin(-infty,0)$.
$x^2over2$ if $xin[0,1)$.
$-x^2+4x-2over2$ if $xin[1,2)$.
$1$ if $xin[2,infty)$
probability uniform-distribution
edited yesterday
Did
246k23221455
asked 2 days ago
Yibei He
1128
1
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday
add a comment |
Let $X_i$ be i.i.d random variables that follow uniform distributions on $[0,1]$. Let $t(x)=min{ngt0:X_1+X_2+...+X_ngt x}$. Find $P(t(x)gt2)$.
Here's my answer but I'm not sure if I interpreted this question correctly.
$$P(t(x)gt2)=1-P(t(x)le2)$$
$$=1-P(n=1 cup n=2)$$
$$=1-P(X_1gt x)-P(X_1+X_2gt xcap X_1lt x)$$
$$=1-1-0=0. if xin(-infty,0)$$
$$=1-int_{x}^{1}dx_1-intint_{x_1+x_2gt x,x_1lt x}dx_1dx_2. if xin[0,1)$$
$$=1-0-intint_{x_1+x_2gt x}dx_1dx_2. ifxin[1,2) $$
$$=1-0-0=1 if xin[2,infty)$$
I did the computation and the answer looks quite "correct".
My answer is
$0$ if $xin(-infty,0)$.
$x^2over2$ if $xin[0,1)$.
$-x^2+4x-2over2$ if $xin[1,2)$.
$1$ if $xin[2,infty)$
probability uniform-distribution
edited yesterday
Did
246k23221455
asked 2 days ago
Yibei He
1128
Let $X_i$ be i.i.d random variables that follow uniform distributions on $[0,1]$. Let $t(x)=min{ngt0:X_1+X_2+...+X_ngt x}$. Find $P(t(x)gt2)$.
Here's my answer but I'm not sure if I interpreted this question correctly.
$$P(t(x)gt2)=1-P(t(x)le2)$$
$$=1-P(n=1 cup n=2)$$
$$=1-P(X_1gt x)-P(X_1+X_2gt xcap X_1lt x)$$
$$=1-1-0=0. if xin(-infty,0)$$
$$=1-int_{x}^{1}dx_1-intint_{x_1+x_2gt x,x_1lt x}dx_1dx_2. if xin[0,1)$$
$$=1-0-intint_{x_1+x_2gt x}dx_1dx_2. ifxin[1,2) $$
$$=1-0-0=1 if xin[2,infty)$$
I did the computation and the answer looks quite "correct".
My answer is
$0$ if $xin(-infty,0)$.
$x^2over2$ if $xin[0,1)$.
$-x^2+4x-2over2$ if $xin[1,2)$.
$1$ if $xin[2,infty)$
probability uniform-distribution
probability uniform-distribution
edited yesterday
Did
246k23221455
asked 2 days ago
Yibei He
1128
edited yesterday
Did
246k23221455
asked 2 days ago
Yibei He
1128
edited yesterday
Did
246k23221455
edited yesterday
Did
246k23221455
edited yesterday
Did
246k23221455
246k23221455
asked 2 days ago
Yibei He
1128
asked 2 days ago
Yibei He
1128
asked 2 days ago
Yibei He
1128
1128
1
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday
add a comment |
1
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday
1
1
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday
add a comment |
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1
Let $T$ be the min index. You can use $${T>2} mbox{ iff } {X_1 + X_2 leq x}$$ Indeed it looks like you are doing this, your integrals and answers look correct.
– Michael
2 days ago
It might be a cleaner derivation if you mentioned the above observation explicitly and/or if you changed your second line to $1-P[{T=1}cup {T=2}]$ rather than $1-P[n=1 cup n=2]$ (since $n$ is not defined and does not look like a random variable).
– Michael
2 days ago
By definition, $P(t(x)>2)$ is the area of the part $x_1+x_2<x$ of the $(x_1,x_2)$-square $[0,1]times[0,1]$. Hence $P(t(x)>2)=frac12x^2$ for $x$ in $(0,1)$ and $P(t(x)>2)=1-frac12(2-x)^2$ for $x$ in $(1,2)$, the other values being obvious.
– Did
yesterday