difference of two independent exponentially distributed random variables
Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.
I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.
Any help would be appreciated, thanks.
probability
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
add a comment |
Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.
I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.
Any help would be appreciated, thanks.
probability
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
add a comment |
Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.
I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.
Any help would be appreciated, thanks.
probability
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
Let $X_1,X_2,X_3$ be independent $Exp(lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.
I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.
Any help would be appreciated, thanks.
probability
probability
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
asked Nov 22 '18 at 17:15
dxdydzdxdydz
1949
1949
add a comment |
add a comment |
1 Answer
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By symmetry,
we have
begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}
Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
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1 Answer
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By symmetry,
we have
begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}
Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
By symmetry,
we have
begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}
Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
By symmetry,
we have
begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}
Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
By symmetry,
we have
begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\&=P(X_2<X_1<X_3)\ &=P(X_2<X_3<X_1)\ &=P(X_3<X_1<X_2)\ &=P(X_3<X_2<X_1)\ end{align}
Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=frac16$$
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 22 '18 at 17:32
Siong Thye GohSiong Thye Goh
99.6k1464117
99.6k1464117
add a comment |
add a comment |
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