Deriving quantiles for $L_1$ using $textbf{P}{ L_n > n cdot a }$ as n tends to infinity
$begingroup$
Given a random variable
$$L_n = sum_{i=1}^n X_i Z_i $$
where $X_i sim Bernoulli(p)$ and $Z_i sim exp(theta)$. I assume ${Z_i }$ is i.i.d. and the sequences ${X_i }$ and ${Z_i }$ are independent.
According to Cramér’s large deviation theorem,
$$lim_{n rightarrow infty} frac{1}{n} log left( textbf{P}{ L_n > n cdot a }right) = - Lambda^*(a)
\
Lambda^*(a) = sup_{xi in mathbb{R}} left( xi a - Lambda (xi) right)
\
Lambda (xi) = log left(kappa(xi) right) $$
where $kappa$ denotes the moment generating function for $X_1 Z_1$ and $a>0$. Note that $kappa(xi) = frac{p theta}{theta - xi}+1-p$. I have calculated the $xi^*$ that solves $Lambda^*(a)$
$$xi(a,p,theta)^*= frac{theta (2-p) - sqrt{p cdot (theta^2p + frac{4(p+theta)}{a}) } }{2(1-p)}.$$
Now, by using the approximation $ textbf{P}{ L_n > n cdot a } approx e^{-n Lambda^*(a)} $, I want to calculate the $alpha$-quantile $(VaR_{alpha})$, where I think of $n$ as "large" and $alpha$, $theta$ and $p$ as given. Formally, I can get this far:
begin{align*}
text{VaR}_alpha(L) = & text{inf} { x in mathbb{R} : P(L>x) leq 1-alpha } \
= & text{inf} { a in mathbb{R} : P left( L>n cdot a right ) leq 1-alpha ) \
approx& inf { a in mathbb{R} : e^{-n Lambda^*(a ) } = 1-alpha
}
end{align*}
where the equality comes from $Lambda^*(a)$ being smooth. Thus I have to find the smallest $a$ that solves $e^{-nLambda^*(a )} = alpha$ for a "large" n.
begin{align*}
e^{-nLambda^*(a )} = 1- alpha
\
e^{-n(xi^* a - log (kappa (xi^*)) )} =1- alpha
\
kappa(xi^*)^n e^{-n xi^* a} = 1- alpha
end{align*}
However, I cannot directly solve this equation since $a$ is in $xi^*$.
How do I proceed?
real-analysis limits probability-theory
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
$endgroup$
add a comment |
$begingroup$
Given a random variable
$$L_n = sum_{i=1}^n X_i Z_i $$
where $X_i sim Bernoulli(p)$ and $Z_i sim exp(theta)$. I assume ${Z_i }$ is i.i.d. and the sequences ${X_i }$ and ${Z_i }$ are independent.
According to Cramér’s large deviation theorem,
$$lim_{n rightarrow infty} frac{1}{n} log left( textbf{P}{ L_n > n cdot a }right) = - Lambda^*(a)
\
Lambda^*(a) = sup_{xi in mathbb{R}} left( xi a - Lambda (xi) right)
\
Lambda (xi) = log left(kappa(xi) right) $$
where $kappa$ denotes the moment generating function for $X_1 Z_1$ and $a>0$. Note that $kappa(xi) = frac{p theta}{theta - xi}+1-p$. I have calculated the $xi^*$ that solves $Lambda^*(a)$
$$xi(a,p,theta)^*= frac{theta (2-p) - sqrt{p cdot (theta^2p + frac{4(p+theta)}{a}) } }{2(1-p)}.$$
Now, by using the approximation $ textbf{P}{ L_n > n cdot a } approx e^{-n Lambda^*(a)} $, I want to calculate the $alpha$-quantile $(VaR_{alpha})$, where I think of $n$ as "large" and $alpha$, $theta$ and $p$ as given. Formally, I can get this far:
begin{align*}
text{VaR}_alpha(L) = & text{inf} { x in mathbb{R} : P(L>x) leq 1-alpha } \
= & text{inf} { a in mathbb{R} : P left( L>n cdot a right ) leq 1-alpha ) \
approx& inf { a in mathbb{R} : e^{-n Lambda^*(a ) } = 1-alpha
}
end{align*}
where the equality comes from $Lambda^*(a)$ being smooth. Thus I have to find the smallest $a$ that solves $e^{-nLambda^*(a )} = alpha$ for a "large" n.
begin{align*}
e^{-nLambda^*(a )} = 1- alpha
\
e^{-n(xi^* a - log (kappa (xi^*)) )} =1- alpha
\
kappa(xi^*)^n e^{-n xi^* a} = 1- alpha
end{align*}
However, I cannot directly solve this equation since $a$ is in $xi^*$.
How do I proceed?
real-analysis limits probability-theory
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
$endgroup$
add a comment |
$begingroup$
Given a random variable
$$L_n = sum_{i=1}^n X_i Z_i $$
where $X_i sim Bernoulli(p)$ and $Z_i sim exp(theta)$. I assume ${Z_i }$ is i.i.d. and the sequences ${X_i }$ and ${Z_i }$ are independent.
According to Cramér’s large deviation theorem,
$$lim_{n rightarrow infty} frac{1}{n} log left( textbf{P}{ L_n > n cdot a }right) = - Lambda^*(a)
\
Lambda^*(a) = sup_{xi in mathbb{R}} left( xi a - Lambda (xi) right)
\
Lambda (xi) = log left(kappa(xi) right) $$
where $kappa$ denotes the moment generating function for $X_1 Z_1$ and $a>0$. Note that $kappa(xi) = frac{p theta}{theta - xi}+1-p$. I have calculated the $xi^*$ that solves $Lambda^*(a)$
$$xi(a,p,theta)^*= frac{theta (2-p) - sqrt{p cdot (theta^2p + frac{4(p+theta)}{a}) } }{2(1-p)}.$$
Now, by using the approximation $ textbf{P}{ L_n > n cdot a } approx e^{-n Lambda^*(a)} $, I want to calculate the $alpha$-quantile $(VaR_{alpha})$, where I think of $n$ as "large" and $alpha$, $theta$ and $p$ as given. Formally, I can get this far:
begin{align*}
text{VaR}_alpha(L) = & text{inf} { x in mathbb{R} : P(L>x) leq 1-alpha } \
= & text{inf} { a in mathbb{R} : P left( L>n cdot a right ) leq 1-alpha ) \
approx& inf { a in mathbb{R} : e^{-n Lambda^*(a ) } = 1-alpha
}
end{align*}
where the equality comes from $Lambda^*(a)$ being smooth. Thus I have to find the smallest $a$ that solves $e^{-nLambda^*(a )} = alpha$ for a "large" n.
begin{align*}
e^{-nLambda^*(a )} = 1- alpha
\
e^{-n(xi^* a - log (kappa (xi^*)) )} =1- alpha
\
kappa(xi^*)^n e^{-n xi^* a} = 1- alpha
end{align*}
However, I cannot directly solve this equation since $a$ is in $xi^*$.
How do I proceed?
real-analysis limits probability-theory
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
$endgroup$
Given a random variable
$$L_n = sum_{i=1}^n X_i Z_i $$
where $X_i sim Bernoulli(p)$ and $Z_i sim exp(theta)$. I assume ${Z_i }$ is i.i.d. and the sequences ${X_i }$ and ${Z_i }$ are independent.
According to Cramér’s large deviation theorem,
$$lim_{n rightarrow infty} frac{1}{n} log left( textbf{P}{ L_n > n cdot a }right) = - Lambda^*(a)
\
Lambda^*(a) = sup_{xi in mathbb{R}} left( xi a - Lambda (xi) right)
\
Lambda (xi) = log left(kappa(xi) right) $$
where $kappa$ denotes the moment generating function for $X_1 Z_1$ and $a>0$. Note that $kappa(xi) = frac{p theta}{theta - xi}+1-p$. I have calculated the $xi^*$ that solves $Lambda^*(a)$
$$xi(a,p,theta)^*= frac{theta (2-p) - sqrt{p cdot (theta^2p + frac{4(p+theta)}{a}) } }{2(1-p)}.$$
Now, by using the approximation $ textbf{P}{ L_n > n cdot a } approx e^{-n Lambda^*(a)} $, I want to calculate the $alpha$-quantile $(VaR_{alpha})$, where I think of $n$ as "large" and $alpha$, $theta$ and $p$ as given. Formally, I can get this far:
begin{align*}
text{VaR}_alpha(L) = & text{inf} { x in mathbb{R} : P(L>x) leq 1-alpha } \
= & text{inf} { a in mathbb{R} : P left( L>n cdot a right ) leq 1-alpha ) \
approx& inf { a in mathbb{R} : e^{-n Lambda^*(a ) } = 1-alpha
}
end{align*}
where the equality comes from $Lambda^*(a)$ being smooth. Thus I have to find the smallest $a$ that solves $e^{-nLambda^*(a )} = alpha$ for a "large" n.
begin{align*}
e^{-nLambda^*(a )} = 1- alpha
\
e^{-n(xi^* a - log (kappa (xi^*)) )} =1- alpha
\
kappa(xi^*)^n e^{-n xi^* a} = 1- alpha
end{align*}
However, I cannot directly solve this equation since $a$ is in $xi^*$.
How do I proceed?
real-analysis limits probability-theory
real-analysis limits probability-theory
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
edited Jan 12 at 15:15
Mads Hulgaard
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
asked Jan 11 at 12:12
Mads HulgaardMads Hulgaard
748
748
add a comment |
add a comment |
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