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?










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?










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?










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






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

2287




2287








  • 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
1






active

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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039629%2fsoa-practice-exam-how-am-i-to-understand-pz-z%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    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









    $endgroup$


















      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




        12.9k11429






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039629%2fsoa-practice-exam-how-am-i-to-understand-pz-z%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix