SOA Practice Exam: How am I to understand P(Z=z)?












2












$begingroup$


Let $X$ denote the loss amount sustained by an insurance company’s policyholder
in an auto collision. Let $Z$ denote the portion of $X$ that the insurance
company will have to pay. An actuary determines that $X$ and $Z$ are independent
with respective density and probability functions
$$
f(x) =
begin{cases}
frac{1}{8} e^{-x/8},& x>0\
0 & mbox{otherwise}
end{cases}
$$



and



$$
P(Z = z) =
begin{cases}
0.45& z = 1, \
0.55&mbox{otherwise}
end{cases}
$$
Calculate the variance of the insurance company’s claim payment $ZX$.



My understanding of $P$ is that there are two events: ${Z=1}$ and ${Z neq 1} = {0 leq Z <1 }$, with probability 0.45 and 0.55 respectively. My confusion arises as I don't see how not knowing the distribution of $Z$ on ${Z neq 1}$ can have no impact on the variance of $ZX$, and a few Mathematica simulations



a = 0;
b = 1;



Variance[Table[ RandomChoice[{0.45, 0.55} -> {1,
RandomVariate[UniformDistribution[{a, b}]]}] RandomVariate[
ExponentialDistribution[1/8]], {100000}]]



seem no confirm this suspicion. However, the answer is supposed to be 44.6. What am I not seeing?






probability variance actuarial-science







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
    $endgroup$
    – Michael
    Dec 14 '18 at 17:11








  • 1




    $begingroup$
    If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
    $endgroup$
    – Michael
    Dec 14 '18 at 17:18








  • 1




    $begingroup$
    Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
    $endgroup$
    – Matthias
    Dec 14 '18 at 17:29
















2












$begingroup$


Let $X$ denote the loss amount sustained by an insurance company’s policyholder
in an auto collision. Let $Z$ denote the portion of $X$ that the insurance
company will have to pay. An actuary determines that $X$ and $Z$ are independent
with respective density and probability functions
$$
f(x) =
begin{cases}
frac{1}{8} e^{-x/8},& x>0\
0 & mbox{otherwise}
end{cases}
$$



and



$$
P(Z = z) =
begin{cases}
0.45& z = 1, \
0.55&mbox{otherwise}
end{cases}
$$
Calculate the variance of the insurance company’s claim payment $ZX$.



My understanding of $P$ is that there are two events: ${Z=1}$ and ${Z neq 1} = {0 leq Z <1 }$, with probability 0.45 and 0.55 respectively. My confusion arises as I don't see how not knowing the distribution of $Z$ on ${Z neq 1}$ can have no impact on the variance of $ZX$, and a few Mathematica simulations



a = 0;
b = 1;



Variance[Table[ RandomChoice[{0.45, 0.55} -> {1,
RandomVariate[UniformDistribution[{a, b}]]}] RandomVariate[
ExponentialDistribution[1/8]], {100000}]]



seem no confirm this suspicion. However, the answer is supposed to be 44.6. What am I not seeing?






probability variance actuarial-science







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
    $endgroup$
    – Michael
    Dec 14 '18 at 17:11








  • 1




    $begingroup$
    If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
    $endgroup$
    – Michael
    Dec 14 '18 at 17:18








  • 1




    $begingroup$
    Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
    $endgroup$
    – Matthias
    Dec 14 '18 at 17:29














2












2








2





$begingroup$


Let $X$ denote the loss amount sustained by an insurance company’s policyholder
in an auto collision. Let $Z$ denote the portion of $X$ that the insurance
company will have to pay. An actuary determines that $X$ and $Z$ are independent
with respective density and probability functions
$$
f(x) =
begin{cases}
frac{1}{8} e^{-x/8},& x>0\
0 & mbox{otherwise}
end{cases}
$$



and



$$
P(Z = z) =
begin{cases}
0.45& z = 1, \
0.55&mbox{otherwise}
end{cases}
$$
Calculate the variance of the insurance company’s claim payment $ZX$.



My understanding of $P$ is that there are two events: ${Z=1}$ and ${Z neq 1} = {0 leq Z <1 }$, with probability 0.45 and 0.55 respectively. My confusion arises as I don't see how not knowing the distribution of $Z$ on ${Z neq 1}$ can have no impact on the variance of $ZX$, and a few Mathematica simulations



a = 0;
b = 1;



Variance[Table[ RandomChoice[{0.45, 0.55} -> {1,
RandomVariate[UniformDistribution[{a, b}]]}] RandomVariate[
ExponentialDistribution[1/8]], {100000}]]



seem no confirm this suspicion. However, the answer is supposed to be 44.6. What am I not seeing?






probability variance actuarial-science







share|cite|improve this question









$endgroup$




Let $X$ denote the loss amount sustained by an insurance company’s policyholder
in an auto collision. Let $Z$ denote the portion of $X$ that the insurance
company will have to pay. An actuary determines that $X$ and $Z$ are independent
with respective density and probability functions
$$
f(x) =
begin{cases}
frac{1}{8} e^{-x/8},& x>0\
0 & mbox{otherwise}
end{cases}
$$



and



$$
P(Z = z) =
begin{cases}
0.45& z = 1, \
0.55&mbox{otherwise}
end{cases}
$$
Calculate the variance of the insurance company’s claim payment $ZX$.



My understanding of $P$ is that there are two events: ${Z=1}$ and ${Z neq 1} = {0 leq Z <1 }$, with probability 0.45 and 0.55 respectively. My confusion arises as I don't see how not knowing the distribution of $Z$ on ${Z neq 1}$ can have no impact on the variance of $ZX$, and a few Mathematica simulations



a = 0;
b = 1;



Variance[Table[ RandomChoice[{0.45, 0.55} -> {1,
RandomVariate[UniformDistribution[{a, b}]]}] RandomVariate[
ExponentialDistribution[1/8]], {100000}]]



seem no confirm this suspicion. However, the answer is supposed to be 44.6. What am I not seeing?






probability variance actuarial-science




probability variance actuarial-science






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 16:59









MatthiasMatthias

3287




3287








  • 1




    $begingroup$
    This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
    $endgroup$
    – Michael
    Dec 14 '18 at 17:11








  • 1




    $begingroup$
    If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
    $endgroup$
    – Michael
    Dec 14 '18 at 17:18








  • 1




    $begingroup$
    Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
    $endgroup$
    – Matthias
    Dec 14 '18 at 17:29














  • 1




    $begingroup$
    This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
    $endgroup$
    – Michael
    Dec 14 '18 at 17:11








  • 1




    $begingroup$
    If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
    $endgroup$
    – Michael
    Dec 14 '18 at 17:18








  • 1




    $begingroup$
    Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
    $endgroup$
    – Matthias
    Dec 14 '18 at 17:29








1




1




$begingroup$
This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
$endgroup$
– Michael
Dec 14 '18 at 17:11






$begingroup$
This is just a typo/mistake in the question itself, they likely meant that $Z$ has one of two possible values, but they forgot to tell you what the other value was. We have $P[Z=1]=0.45$ and $P[Z=theta] = 0.55$ but we do not know the value of $theta$, because the problem forgot to tell us that.
$endgroup$
– Michael
Dec 14 '18 at 17:11






1




1




$begingroup$
If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
$endgroup$
– Michael
Dec 14 '18 at 17:18






$begingroup$
If you assume $theta=0$ then you can get the outcome of $Var(XZ)=44.64$, which seems consistent with the number you say you are supposed to get. (It also makes intuitive sense: Either the company pays the full amount, or 0.)
$endgroup$
– Michael
Dec 14 '18 at 17:18






1




1




$begingroup$
Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
$endgroup$
– Matthias
Dec 14 '18 at 17:29




$begingroup$
Confirmed 44.64 for $theta = 0$. Copy-and-paste error by the problem author, it seems. Thanks!
$endgroup$
– Matthias
Dec 14 '18 at 17:29










1 Answer
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oldest

votes


















0












$begingroup$

Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been:
begin{align*}
P[Z=1]&=0.45\
P[Z=0]&=0.55
end{align*}
from that you can indeed show $Var(XZ)=44.64$.






share|cite|improve this answer









$endgroup$













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    0












    $begingroup$

    Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been:
    begin{align*}
    P[Z=1]&=0.45\
    P[Z=0]&=0.55
    end{align*}
    from that you can indeed show $Var(XZ)=44.64$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been:
      begin{align*}
      P[Z=1]&=0.45\
      P[Z=0]&=0.55
      end{align*}
      from that you can indeed show $Var(XZ)=44.64$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been:
        begin{align*}
        P[Z=1]&=0.45\
        P[Z=0]&=0.55
        end{align*}
        from that you can indeed show $Var(XZ)=44.64$.






        share|cite|improve this answer









        $endgroup$



        Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been:
        begin{align*}
        P[Z=1]&=0.45\
        P[Z=0]&=0.55
        end{align*}
        from that you can indeed show $Var(XZ)=44.64$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 17:32









        MichaelMichael

        12.9k11429




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