Proof of Cramér-Lundberg inequalityChecking proof that a given process is a martingaleExercise on martingale and sub/supermartingaleExponential Integrals of Brownian Motion, Stopping Times, Expectation$L^2$-bound for the stochastic integralIf $M$ is a local martingale and $τ:=infleft≥εright$, then $text P[[M]_∞≥δ]≤text P[τ<∞]+text P[[M]_τ≥δ]$Normal random variable and martingaleDurrett: Showing $E(log Y)$ is negative when $Y$ is nonnegative with $E(Y) = 1$ and $P(Y = 1) < 1$Strictly Convex function (Help with Durrett's exercise)Compound Poisson models with completely monotone claim sizesConvergence using Laplace transform
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Proof of Cramér-Lundberg inequality
Checking proof that a given process is a martingaleExercise on martingale and sub/supermartingaleExponential Integrals of Brownian Motion, Stopping Times, Expectation$L^2$-bound for the stochastic integralIf $M$ is a local martingale and $τ:=infleftM_t$, then $text P[[M]_∞≥δ]≤text P[τ<∞]+text P[[M]_τ≥δ]$Normal random variable and martingaleDurrett: Showing $E(log Y)$ is negative when $Y$ is nonnegative with $E(Y) = 1$ and $P(Y = 1) < 1$Strictly Convex function (Help with Durrett's exercise)Compound Poisson models with completely monotone claim sizesConvergence using Laplace transform
$begingroup$
I'm trying to prove the Cramér-Lundberg inequality, which deals with the probability of ruin for an insurance company given a certain initial capital. Specifically, if $Y_1, Y_2, ldots$ are the differences between the premiums and payments of an insurance company at time $n$, and $X_n = Y_1 + cdots + Y_n$ is the total gain of the insurance company at time $n$, and $k_0$ is the initial capital, then the probability of eventual ruin $p(k_0)$ satisfies the Cramér-Lundberg inequality:
$$
p(k_0) := mathbb Pleft[ infleft X_n + k_0 : n in mathbb N_0 right < 0 right] leq expleft(theta^* k_0right)
$$
where $theta^* < 0$ satisfies $logleft( mathbb Eleft[ exp(theta^* Y_1 )right]right) = 0$.
My reference text proposes proving this in the following steps. Suppose $Y_1, Y_2, ldots$ are i.i.d. integrable random variables that are not almost surely constant. Let $X_n = Y_1 + cdots + Y_n$, and suppose there is $delta > 0$ so that $mathbb Eleft[expleft(theta Y_1 right)right] < infty$ for all $theta in (-delta, delta)$. Define $psi : (-delta, delta) to mathbb R$ by $$psi(theta) := log left(mathbb Eleft[expleft(theta Y_1 right)right]right)$$
and define the process $Z^theta = left(Z^theta_nright)_n geq 1$ by $Z_n^theta := expleft(theta X_n - npsi(theta)right)$. We are suggested to show the following:
$Z^theta$ is a martingale for all $theta in (-delta, delta)$.
$psi$ is strictly convex.
$mathbb Eleft[sqrtZ_n^thetaright] xrightarrown to infty 0$ for $theta neq 0$.
$Z_n^theta xrightarrown to infty 0$ almost surely.- If $psi(theta) = 0$ has a nonzero solution $theta^*$, then $theta^* < 0$.
- Prove that if such a $theta^*$ exists, then $p(k_0) leq expleft(theta^* k_0right)$.
I've been able to show 1, 4 (from 3), and 5 (from 2). I'm close for 2 but having some trouble: we need to show $psi(lambdatheta + (1-lambda)phi) < lambda psi(theta) + (1-lambda)psi(phi)$ whenever $theta neq phi$ and $lambda in (0,1)$. Clearly we have $$psi(lambdatheta + (1-lambda)phi) = logmathbb Eleft[ expleft(lambdatheta Y_1 + (1-lambda)phi Y_1 right)right]$$ Meanwhile by Jensen's inequality and concavity of $x mapsto x^lambda$ for $0 < lambda < 1$,
beginalign*
lambda psi(theta) + (1-lambda)psi(phi) &= log left( mathbb Eleft[exp (theta Y_1) right]^lambdaright) + logleft(mathbb Eleft[ exp(phi Y_1)right]^1-lambdaright) \
&geq logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]right) + logleft(mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right) \
&= logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right).
endalign*
If I could show $mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right] geq mathbb Eleft[expleft(lambda theta Y_1right)expleft((1-lambda)phi Y_1right)right]$,that would solve this problem, but this is very far from obvious to me (especially since the integrands aren't independent).
Then 3 and 6 I'm really stuck on. Any help on any of these three would be greatly appreciated. Note I would prefer not to use martingale convergence theorems because these results have yet to appear in my textbook; I can only work with square integrable martingales and stopping times.
real-analysis probability-theory stochastic-processes martingales stopping-times
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the Cramér-Lundberg inequality, which deals with the probability of ruin for an insurance company given a certain initial capital. Specifically, if $Y_1, Y_2, ldots$ are the differences between the premiums and payments of an insurance company at time $n$, and $X_n = Y_1 + cdots + Y_n$ is the total gain of the insurance company at time $n$, and $k_0$ is the initial capital, then the probability of eventual ruin $p(k_0)$ satisfies the Cramér-Lundberg inequality:
$$
p(k_0) := mathbb Pleft[ infleft X_n + k_0 : n in mathbb N_0 right < 0 right] leq expleft(theta^* k_0right)
$$
where $theta^* < 0$ satisfies $logleft( mathbb Eleft[ exp(theta^* Y_1 )right]right) = 0$.
My reference text proposes proving this in the following steps. Suppose $Y_1, Y_2, ldots$ are i.i.d. integrable random variables that are not almost surely constant. Let $X_n = Y_1 + cdots + Y_n$, and suppose there is $delta > 0$ so that $mathbb Eleft[expleft(theta Y_1 right)right] < infty$ for all $theta in (-delta, delta)$. Define $psi : (-delta, delta) to mathbb R$ by $$psi(theta) := log left(mathbb Eleft[expleft(theta Y_1 right)right]right)$$
and define the process $Z^theta = left(Z^theta_nright)_n geq 1$ by $Z_n^theta := expleft(theta X_n - npsi(theta)right)$. We are suggested to show the following:
$Z^theta$ is a martingale for all $theta in (-delta, delta)$.
$psi$ is strictly convex.
$mathbb Eleft[sqrtZ_n^thetaright] xrightarrown to infty 0$ for $theta neq 0$.
$Z_n^theta xrightarrown to infty 0$ almost surely.- If $psi(theta) = 0$ has a nonzero solution $theta^*$, then $theta^* < 0$.
- Prove that if such a $theta^*$ exists, then $p(k_0) leq expleft(theta^* k_0right)$.
I've been able to show 1, 4 (from 3), and 5 (from 2). I'm close for 2 but having some trouble: we need to show $psi(lambdatheta + (1-lambda)phi) < lambda psi(theta) + (1-lambda)psi(phi)$ whenever $theta neq phi$ and $lambda in (0,1)$. Clearly we have $$psi(lambdatheta + (1-lambda)phi) = logmathbb Eleft[ expleft(lambdatheta Y_1 + (1-lambda)phi Y_1 right)right]$$ Meanwhile by Jensen's inequality and concavity of $x mapsto x^lambda$ for $0 < lambda < 1$,
beginalign*
lambda psi(theta) + (1-lambda)psi(phi) &= log left( mathbb Eleft[exp (theta Y_1) right]^lambdaright) + logleft(mathbb Eleft[ exp(phi Y_1)right]^1-lambdaright) \
&geq logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]right) + logleft(mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right) \
&= logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right).
endalign*
If I could show $mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right] geq mathbb Eleft[expleft(lambda theta Y_1right)expleft((1-lambda)phi Y_1right)right]$,that would solve this problem, but this is very far from obvious to me (especially since the integrands aren't independent).
Then 3 and 6 I'm really stuck on. Any help on any of these three would be greatly appreciated. Note I would prefer not to use martingale convergence theorems because these results have yet to appear in my textbook; I can only work with square integrable martingales and stopping times.
real-analysis probability-theory stochastic-processes martingales stopping-times
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the Cramér-Lundberg inequality, which deals with the probability of ruin for an insurance company given a certain initial capital. Specifically, if $Y_1, Y_2, ldots$ are the differences between the premiums and payments of an insurance company at time $n$, and $X_n = Y_1 + cdots + Y_n$ is the total gain of the insurance company at time $n$, and $k_0$ is the initial capital, then the probability of eventual ruin $p(k_0)$ satisfies the Cramér-Lundberg inequality:
$$
p(k_0) := mathbb Pleft[ infleft X_n + k_0 : n in mathbb N_0 right < 0 right] leq expleft(theta^* k_0right)
$$
where $theta^* < 0$ satisfies $logleft( mathbb Eleft[ exp(theta^* Y_1 )right]right) = 0$.
My reference text proposes proving this in the following steps. Suppose $Y_1, Y_2, ldots$ are i.i.d. integrable random variables that are not almost surely constant. Let $X_n = Y_1 + cdots + Y_n$, and suppose there is $delta > 0$ so that $mathbb Eleft[expleft(theta Y_1 right)right] < infty$ for all $theta in (-delta, delta)$. Define $psi : (-delta, delta) to mathbb R$ by $$psi(theta) := log left(mathbb Eleft[expleft(theta Y_1 right)right]right)$$
and define the process $Z^theta = left(Z^theta_nright)_n geq 1$ by $Z_n^theta := expleft(theta X_n - npsi(theta)right)$. We are suggested to show the following:
$Z^theta$ is a martingale for all $theta in (-delta, delta)$.
$psi$ is strictly convex.
$mathbb Eleft[sqrtZ_n^thetaright] xrightarrown to infty 0$ for $theta neq 0$.
$Z_n^theta xrightarrown to infty 0$ almost surely.- If $psi(theta) = 0$ has a nonzero solution $theta^*$, then $theta^* < 0$.
- Prove that if such a $theta^*$ exists, then $p(k_0) leq expleft(theta^* k_0right)$.
I've been able to show 1, 4 (from 3), and 5 (from 2). I'm close for 2 but having some trouble: we need to show $psi(lambdatheta + (1-lambda)phi) < lambda psi(theta) + (1-lambda)psi(phi)$ whenever $theta neq phi$ and $lambda in (0,1)$. Clearly we have $$psi(lambdatheta + (1-lambda)phi) = logmathbb Eleft[ expleft(lambdatheta Y_1 + (1-lambda)phi Y_1 right)right]$$ Meanwhile by Jensen's inequality and concavity of $x mapsto x^lambda$ for $0 < lambda < 1$,
beginalign*
lambda psi(theta) + (1-lambda)psi(phi) &= log left( mathbb Eleft[exp (theta Y_1) right]^lambdaright) + logleft(mathbb Eleft[ exp(phi Y_1)right]^1-lambdaright) \
&geq logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]right) + logleft(mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right) \
&= logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right).
endalign*
If I could show $mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right] geq mathbb Eleft[expleft(lambda theta Y_1right)expleft((1-lambda)phi Y_1right)right]$,that would solve this problem, but this is very far from obvious to me (especially since the integrands aren't independent).
Then 3 and 6 I'm really stuck on. Any help on any of these three would be greatly appreciated. Note I would prefer not to use martingale convergence theorems because these results have yet to appear in my textbook; I can only work with square integrable martingales and stopping times.
real-analysis probability-theory stochastic-processes martingales stopping-times
$endgroup$
I'm trying to prove the Cramér-Lundberg inequality, which deals with the probability of ruin for an insurance company given a certain initial capital. Specifically, if $Y_1, Y_2, ldots$ are the differences between the premiums and payments of an insurance company at time $n$, and $X_n = Y_1 + cdots + Y_n$ is the total gain of the insurance company at time $n$, and $k_0$ is the initial capital, then the probability of eventual ruin $p(k_0)$ satisfies the Cramér-Lundberg inequality:
$$
p(k_0) := mathbb Pleft[ infleft X_n + k_0 : n in mathbb N_0 right < 0 right] leq expleft(theta^* k_0right)
$$
where $theta^* < 0$ satisfies $logleft( mathbb Eleft[ exp(theta^* Y_1 )right]right) = 0$.
My reference text proposes proving this in the following steps. Suppose $Y_1, Y_2, ldots$ are i.i.d. integrable random variables that are not almost surely constant. Let $X_n = Y_1 + cdots + Y_n$, and suppose there is $delta > 0$ so that $mathbb Eleft[expleft(theta Y_1 right)right] < infty$ for all $theta in (-delta, delta)$. Define $psi : (-delta, delta) to mathbb R$ by $$psi(theta) := log left(mathbb Eleft[expleft(theta Y_1 right)right]right)$$
and define the process $Z^theta = left(Z^theta_nright)_n geq 1$ by $Z_n^theta := expleft(theta X_n - npsi(theta)right)$. We are suggested to show the following:
$Z^theta$ is a martingale for all $theta in (-delta, delta)$.
$psi$ is strictly convex.
$mathbb Eleft[sqrtZ_n^thetaright] xrightarrown to infty 0$ for $theta neq 0$.
$Z_n^theta xrightarrown to infty 0$ almost surely.- If $psi(theta) = 0$ has a nonzero solution $theta^*$, then $theta^* < 0$.
- Prove that if such a $theta^*$ exists, then $p(k_0) leq expleft(theta^* k_0right)$.
I've been able to show 1, 4 (from 3), and 5 (from 2). I'm close for 2 but having some trouble: we need to show $psi(lambdatheta + (1-lambda)phi) < lambda psi(theta) + (1-lambda)psi(phi)$ whenever $theta neq phi$ and $lambda in (0,1)$. Clearly we have $$psi(lambdatheta + (1-lambda)phi) = logmathbb Eleft[ expleft(lambdatheta Y_1 + (1-lambda)phi Y_1 right)right]$$ Meanwhile by Jensen's inequality and concavity of $x mapsto x^lambda$ for $0 < lambda < 1$,
beginalign*
lambda psi(theta) + (1-lambda)psi(phi) &= log left( mathbb Eleft[exp (theta Y_1) right]^lambdaright) + logleft(mathbb Eleft[ exp(phi Y_1)right]^1-lambdaright) \
&geq logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]right) + logleft(mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right) \
&= logleft(mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right]right).
endalign*
If I could show $mathbb Eleft[expleft(lambda theta Y_1right)right]mathbb Eleft[expleft((1-lambda)phi Y_1right)right] geq mathbb Eleft[expleft(lambda theta Y_1right)expleft((1-lambda)phi Y_1right)right]$,that would solve this problem, but this is very far from obvious to me (especially since the integrands aren't independent).
Then 3 and 6 I'm really stuck on. Any help on any of these three would be greatly appreciated. Note I would prefer not to use martingale convergence theorems because these results have yet to appear in my textbook; I can only work with square integrable martingales and stopping times.
real-analysis probability-theory stochastic-processes martingales stopping-times
real-analysis probability-theory stochastic-processes martingales stopping-times
asked Mar 29 at 3:45
D FordD Ford
675313
675313
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$begingroup$
Here is an idea for the convergence results 3. and 4.
I will start with 4. : Write $Z_n^theta=e^nbig(thetafracX_nn-psi(theta)big)$ and let $A^theta_n=thetafracX_nn-psi(theta);,ninmathbbN$. By the strong law of large numbers, it follows that for all $thetain(-delta,delta)$ $$A^theta_nundersetntoinftylongrightarrow a(theta):=mathbbE[theta ;Y_1]-psi(theta);;texta.s.$$ Now, observe that for all $thetaneq0$
$$a(theta)= mathbbE[theta;Y_1]-log mathbbE[exptheta;Y_1]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $tildeOmega$ of probability $1$, there exists an $n_0inmathbbN$ and $M<0$, such that for all $ngeq n_0$,
$$A^theta_nleq M;;(on;tildeOmega);$$
Hence for $thetaneq 0$ $$ |Z^theta_n|leq e^nMlongrightarrow 0;;a.s.$$ (For $theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $sqrtZ_n^theta$ converges to $0$ almost surely (provided that $thetaneq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
$endgroup$
add a comment |
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$begingroup$
Here is an idea for the convergence results 3. and 4.
I will start with 4. : Write $Z_n^theta=e^nbig(thetafracX_nn-psi(theta)big)$ and let $A^theta_n=thetafracX_nn-psi(theta);,ninmathbbN$. By the strong law of large numbers, it follows that for all $thetain(-delta,delta)$ $$A^theta_nundersetntoinftylongrightarrow a(theta):=mathbbE[theta ;Y_1]-psi(theta);;texta.s.$$ Now, observe that for all $thetaneq0$
$$a(theta)= mathbbE[theta;Y_1]-log mathbbE[exptheta;Y_1]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $tildeOmega$ of probability $1$, there exists an $n_0inmathbbN$ and $M<0$, such that for all $ngeq n_0$,
$$A^theta_nleq M;;(on;tildeOmega);$$
Hence for $thetaneq 0$ $$ |Z^theta_n|leq e^nMlongrightarrow 0;;a.s.$$ (For $theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $sqrtZ_n^theta$ converges to $0$ almost surely (provided that $thetaneq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
$endgroup$
add a comment |
$begingroup$
Here is an idea for the convergence results 3. and 4.
I will start with 4. : Write $Z_n^theta=e^nbig(thetafracX_nn-psi(theta)big)$ and let $A^theta_n=thetafracX_nn-psi(theta);,ninmathbbN$. By the strong law of large numbers, it follows that for all $thetain(-delta,delta)$ $$A^theta_nundersetntoinftylongrightarrow a(theta):=mathbbE[theta ;Y_1]-psi(theta);;texta.s.$$ Now, observe that for all $thetaneq0$
$$a(theta)= mathbbE[theta;Y_1]-log mathbbE[exptheta;Y_1]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $tildeOmega$ of probability $1$, there exists an $n_0inmathbbN$ and $M<0$, such that for all $ngeq n_0$,
$$A^theta_nleq M;;(on;tildeOmega);$$
Hence for $thetaneq 0$ $$ |Z^theta_n|leq e^nMlongrightarrow 0;;a.s.$$ (For $theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $sqrtZ_n^theta$ converges to $0$ almost surely (provided that $thetaneq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
$endgroup$
add a comment |
$begingroup$
Here is an idea for the convergence results 3. and 4.
I will start with 4. : Write $Z_n^theta=e^nbig(thetafracX_nn-psi(theta)big)$ and let $A^theta_n=thetafracX_nn-psi(theta);,ninmathbbN$. By the strong law of large numbers, it follows that for all $thetain(-delta,delta)$ $$A^theta_nundersetntoinftylongrightarrow a(theta):=mathbbE[theta ;Y_1]-psi(theta);;texta.s.$$ Now, observe that for all $thetaneq0$
$$a(theta)= mathbbE[theta;Y_1]-log mathbbE[exptheta;Y_1]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $tildeOmega$ of probability $1$, there exists an $n_0inmathbbN$ and $M<0$, such that for all $ngeq n_0$,
$$A^theta_nleq M;;(on;tildeOmega);$$
Hence for $thetaneq 0$ $$ |Z^theta_n|leq e^nMlongrightarrow 0;;a.s.$$ (For $theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $sqrtZ_n^theta$ converges to $0$ almost surely (provided that $thetaneq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
$endgroup$
Here is an idea for the convergence results 3. and 4.
I will start with 4. : Write $Z_n^theta=e^nbig(thetafracX_nn-psi(theta)big)$ and let $A^theta_n=thetafracX_nn-psi(theta);,ninmathbbN$. By the strong law of large numbers, it follows that for all $thetain(-delta,delta)$ $$A^theta_nundersetntoinftylongrightarrow a(theta):=mathbbE[theta ;Y_1]-psi(theta);;texta.s.$$ Now, observe that for all $thetaneq0$
$$a(theta)= mathbbE[theta;Y_1]-log mathbbE[exptheta;Y_1]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $tildeOmega$ of probability $1$, there exists an $n_0inmathbbN$ and $M<0$, such that for all $ngeq n_0$,
$$A^theta_nleq M;;(on;tildeOmega);$$
Hence for $thetaneq 0$ $$ |Z^theta_n|leq e^nMlongrightarrow 0;;a.s.$$ (For $theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $sqrtZ_n^theta$ converges to $0$ almost surely (provided that $thetaneq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
answered Mar 29 at 8:29
gigastergigaster
1413
1413
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