Create combined CDF (Value at Risk)
this is my first post and I have probably already made a fool out of myself by the title but I'm desperate to understand this. My question comes from an example so I have the solution but there is a step I want to understand the theory behind.
Q: $L$ describes percentual damage to a building within next year. We know that $P(L=0)=0.95$. With $p=0.05$ a damage happens and then $L$ has uniform distribution on [50,100]. Find smallest value of m such that $P(L>m)leq 0.001$
A:
We know that $P(L=0)=0.95$ and $P(50leq L leq 100)=0.05$ and:
begin{equation}
P(L>m)=1-P(Lleq m)leq 0.001 rightarrow P(Lleq m) geq 0.999
end{equation}
And we know that a CDF is defined by $F_L(m)=P(Lleq m)$ and we get $F_L$ as:
begin{equation}
F_L(m) = begin{cases} 0 & m<0 \ 0.95 & 0 leq m <50 \ 0.95 + 0.05 cdot frac{m-50}{50} & 50 leq m leq 100 \1 & m>100end{cases}
end{equation}
My Question: What is the theory behind geting $0.95 + 0.05 cdot frac{m-50}{50}$ especially the second term. I do know that the CDF for a uniform r.v. has the form of $frac{x-a}{b-a}$ but I don't understand how we combine the cases of no damage and damage and then multiply the probability 0.05 with the CDF for a uniform r.v. I do though understand why we add the 0.95.
I have solved for m and that gives m=99 but there is some kind of essential theory that I don't get. Maybe if I know the correct term I could google it but as my title gives away I'm lost.
probability probability-theory
asked Nov 28 at 18:47
probablyme
133
add a comment |
this is my first post and I have probably already made a fool out of myself by the title but I'm desperate to understand this. My question comes from an example so I have the solution but there is a step I want to understand the theory behind.
Q: $L$ describes percentual damage to a building within next year. We know that $P(L=0)=0.95$. With $p=0.05$ a damage happens and then $L$ has uniform distribution on [50,100]. Find smallest value of m such that $P(L>m)leq 0.001$
A:
We know that $P(L=0)=0.95$ and $P(50leq L leq 100)=0.05$ and:
begin{equation}
P(L>m)=1-P(Lleq m)leq 0.001 rightarrow P(Lleq m) geq 0.999
end{equation}
And we know that a CDF is defined by $F_L(m)=P(Lleq m)$ and we get $F_L$ as:
begin{equation}
F_L(m) = begin{cases} 0 & m<0 \ 0.95 & 0 leq m <50 \ 0.95 + 0.05 cdot frac{m-50}{50} & 50 leq m leq 100 \1 & m>100end{cases}
end{equation}
My Question: What is the theory behind geting $0.95 + 0.05 cdot frac{m-50}{50}$ especially the second term. I do know that the CDF for a uniform r.v. has the form of $frac{x-a}{b-a}$ but I don't understand how we combine the cases of no damage and damage and then multiply the probability 0.05 with the CDF for a uniform r.v. I do though understand why we add the 0.95.
I have solved for m and that gives m=99 but there is some kind of essential theory that I don't get. Maybe if I know the correct term I could google it but as my title gives away I'm lost.
probability probability-theory
asked Nov 28 at 18:47
probablyme
133
add a comment |
this is my first post and I have probably already made a fool out of myself by the title but I'm desperate to understand this. My question comes from an example so I have the solution but there is a step I want to understand the theory behind.
Q: $L$ describes percentual damage to a building within next year. We know that $P(L=0)=0.95$. With $p=0.05$ a damage happens and then $L$ has uniform distribution on [50,100]. Find smallest value of m such that $P(L>m)leq 0.001$
A:
We know that $P(L=0)=0.95$ and $P(50leq L leq 100)=0.05$ and:
begin{equation}
P(L>m)=1-P(Lleq m)leq 0.001 rightarrow P(Lleq m) geq 0.999
end{equation}
And we know that a CDF is defined by $F_L(m)=P(Lleq m)$ and we get $F_L$ as:
begin{equation}
F_L(m) = begin{cases} 0 & m<0 \ 0.95 & 0 leq m <50 \ 0.95 + 0.05 cdot frac{m-50}{50} & 50 leq m leq 100 \1 & m>100end{cases}
end{equation}
My Question: What is the theory behind geting $0.95 + 0.05 cdot frac{m-50}{50}$ especially the second term. I do know that the CDF for a uniform r.v. has the form of $frac{x-a}{b-a}$ but I don't understand how we combine the cases of no damage and damage and then multiply the probability 0.05 with the CDF for a uniform r.v. I do though understand why we add the 0.95.
I have solved for m and that gives m=99 but there is some kind of essential theory that I don't get. Maybe if I know the correct term I could google it but as my title gives away I'm lost.
probability probability-theory
asked Nov 28 at 18:47
probablyme
133
this is my first post and I have probably already made a fool out of myself by the title but I'm desperate to understand this. My question comes from an example so I have the solution but there is a step I want to understand the theory behind.
Q: $L$ describes percentual damage to a building within next year. We know that $P(L=0)=0.95$. With $p=0.05$ a damage happens and then $L$ has uniform distribution on [50,100]. Find smallest value of m such that $P(L>m)leq 0.001$
A:
We know that $P(L=0)=0.95$ and $P(50leq L leq 100)=0.05$ and:
begin{equation}
P(L>m)=1-P(Lleq m)leq 0.001 rightarrow P(Lleq m) geq 0.999
end{equation}
And we know that a CDF is defined by $F_L(m)=P(Lleq m)$ and we get $F_L$ as:
begin{equation}
F_L(m) = begin{cases} 0 & m<0 \ 0.95 & 0 leq m <50 \ 0.95 + 0.05 cdot frac{m-50}{50} & 50 leq m leq 100 \1 & m>100end{cases}
end{equation}
My Question: What is the theory behind geting $0.95 + 0.05 cdot frac{m-50}{50}$ especially the second term. I do know that the CDF for a uniform r.v. has the form of $frac{x-a}{b-a}$ but I don't understand how we combine the cases of no damage and damage and then multiply the probability 0.05 with the CDF for a uniform r.v. I do though understand why we add the 0.95.
I have solved for m and that gives m=99 but there is some kind of essential theory that I don't get. Maybe if I know the correct term I could google it but as my title gives away I'm lost.
probability probability-theory
probability probability-theory
asked Nov 28 at 18:47
probablyme
133
asked Nov 28 at 18:47
probablyme
133
asked Nov 28 at 18:47
probablyme
133
asked Nov 28 at 18:47
probablyme
133
asked Nov 28 at 18:47
probablyme
133
133
add a comment |
add a comment |
1 Answer
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Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(Lleq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50leq mleq 100$,
$$P(Lleq m)=P(L< 50)+P(50leq Lleq m) = .95 + .05cdotfrac{m-50}{50}$$
answered Nov 28 at 19:35
pwerth
1,500411
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
add a comment |
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1 Answer
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Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(Lleq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50leq mleq 100$,
$$P(Lleq m)=P(L< 50)+P(50leq Lleq m) = .95 + .05cdotfrac{m-50}{50}$$
answered Nov 28 at 19:35
pwerth
1,500411
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
add a comment |
Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(Lleq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50leq mleq 100$,
$$P(Lleq m)=P(L< 50)+P(50leq Lleq m) = .95 + .05cdotfrac{m-50}{50}$$
answered Nov 28 at 19:35
pwerth
1,500411
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
add a comment |
Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(Lleq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50leq mleq 100$,
$$P(Lleq m)=P(L< 50)+P(50leq Lleq m) = .95 + .05cdotfrac{m-50}{50}$$
answered Nov 28 at 19:35
pwerth
1,500411
Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(Lleq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50leq mleq 100$,
$$P(Lleq m)=P(L< 50)+P(50leq Lleq m) = .95 + .05cdotfrac{m-50}{50}$$
answered Nov 28 at 19:35
pwerth
1,500411
answered Nov 28 at 19:35
pwerth
1,500411
answered Nov 28 at 19:35
pwerth
1,500411
answered Nov 28 at 19:35
pwerth
1,500411
1,500411
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
add a comment |
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
Thank you very much for the great answer to my question. Would it be correct to say that $F_L (m)$ is a combination of a discrete and continuous r.v.? Where I think of the first part as a discrete r.v. And the uniform as a continuous.
– probablyme
Nov 28 at 19:44
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
@probablyme Be careful - $F_{L}(m)$ is not a r.v., that would be $L$. What exactly do you mean by "combination"? I suppose you could write $L=.95cdot 1_{{no damage}} + .05cdot 1_{{damage}}cdot U,$ where $U$ is uniform on $[50,100]$.
– pwerth
Nov 28 at 19:52
add a comment |
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