If $P(X = c) > 0$ show that $F_{X}$ is discontinuous at $c$












1












$begingroup$


Let $(Omega, mathcal{F}, mathbb P)$ and $c in mathbb R$ so that $mathbb P(X = c) > 0$



Show that the distribution function $F_{X}$ of $X$ has a point of discontinuity at $c$.



My ideas:



Since $F_{X}$ is left-continuous by definition we need to look at the right-hand side. Let $epsilon > 0$



Note that $F_{X}(c+ epsilon)=mathbb P(X leq c + epsilon)=mathbb P(X<c+epsilon)+mathbb P(X=c+epsilon)$ and then I want to introduce $delta in (0, epsilon)$ so that $mathbb P(X leq c + epsilon)geqmathbb P(X<c+delta)+mathbb P(X=c+epsilon)$ and then letting $epsilon to 0$ we have
$P(X<c+delta)+mathbb P(X=c+epsilon)to P(X < c + delta)+P(X=c) >P(X < c + delta)geq P(Xleq c)$



I am unsure about the above, because since I have let $epsilon to 0$, does that mean by definition of $delta$ that $delta to 0$?



I am also struggling to find a case where the converse does not hold, i.e. if $F_{X}$ has a point of discontinuity at $c$ then $mathbb P (X=c)leq0$.






probability-theory probability-distributions discontinuous-functions







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  • 1




    $begingroup$
    $F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
    $endgroup$
    – Will M.
    Dec 2 '18 at 22:27
















1












$begingroup$


Let $(Omega, mathcal{F}, mathbb P)$ and $c in mathbb R$ so that $mathbb P(X = c) > 0$



Show that the distribution function $F_{X}$ of $X$ has a point of discontinuity at $c$.



My ideas:



Since $F_{X}$ is left-continuous by definition we need to look at the right-hand side. Let $epsilon > 0$



Note that $F_{X}(c+ epsilon)=mathbb P(X leq c + epsilon)=mathbb P(X<c+epsilon)+mathbb P(X=c+epsilon)$ and then I want to introduce $delta in (0, epsilon)$ so that $mathbb P(X leq c + epsilon)geqmathbb P(X<c+delta)+mathbb P(X=c+epsilon)$ and then letting $epsilon to 0$ we have
$P(X<c+delta)+mathbb P(X=c+epsilon)to P(X < c + delta)+P(X=c) >P(X < c + delta)geq P(Xleq c)$



I am unsure about the above, because since I have let $epsilon to 0$, does that mean by definition of $delta$ that $delta to 0$?



I am also struggling to find a case where the converse does not hold, i.e. if $F_{X}$ has a point of discontinuity at $c$ then $mathbb P (X=c)leq0$.






probability-theory probability-distributions discontinuous-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
    $endgroup$
    – Will M.
    Dec 2 '18 at 22:27














1












1








1





$begingroup$


Let $(Omega, mathcal{F}, mathbb P)$ and $c in mathbb R$ so that $mathbb P(X = c) > 0$



Show that the distribution function $F_{X}$ of $X$ has a point of discontinuity at $c$.



My ideas:



Since $F_{X}$ is left-continuous by definition we need to look at the right-hand side. Let $epsilon > 0$



Note that $F_{X}(c+ epsilon)=mathbb P(X leq c + epsilon)=mathbb P(X<c+epsilon)+mathbb P(X=c+epsilon)$ and then I want to introduce $delta in (0, epsilon)$ so that $mathbb P(X leq c + epsilon)geqmathbb P(X<c+delta)+mathbb P(X=c+epsilon)$ and then letting $epsilon to 0$ we have
$P(X<c+delta)+mathbb P(X=c+epsilon)to P(X < c + delta)+P(X=c) >P(X < c + delta)geq P(Xleq c)$



I am unsure about the above, because since I have let $epsilon to 0$, does that mean by definition of $delta$ that $delta to 0$?



I am also struggling to find a case where the converse does not hold, i.e. if $F_{X}$ has a point of discontinuity at $c$ then $mathbb P (X=c)leq0$.






probability-theory probability-distributions discontinuous-functions







share|cite|improve this question









$endgroup$




Let $(Omega, mathcal{F}, mathbb P)$ and $c in mathbb R$ so that $mathbb P(X = c) > 0$



Show that the distribution function $F_{X}$ of $X$ has a point of discontinuity at $c$.



My ideas:



Since $F_{X}$ is left-continuous by definition we need to look at the right-hand side. Let $epsilon > 0$



Note that $F_{X}(c+ epsilon)=mathbb P(X leq c + epsilon)=mathbb P(X<c+epsilon)+mathbb P(X=c+epsilon)$ and then I want to introduce $delta in (0, epsilon)$ so that $mathbb P(X leq c + epsilon)geqmathbb P(X<c+delta)+mathbb P(X=c+epsilon)$ and then letting $epsilon to 0$ we have
$P(X<c+delta)+mathbb P(X=c+epsilon)to P(X < c + delta)+P(X=c) >P(X < c + delta)geq P(Xleq c)$



I am unsure about the above, because since I have let $epsilon to 0$, does that mean by definition of $delta$ that $delta to 0$?



I am also struggling to find a case where the converse does not hold, i.e. if $F_{X}$ has a point of discontinuity at $c$ then $mathbb P (X=c)leq0$.






probability-theory probability-distributions discontinuous-functions




probability-theory probability-distributions discontinuous-functions






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asked Dec 2 '18 at 22:04









SABOYSABOY

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  • 1




    $begingroup$
    $F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
    $endgroup$
    – Will M.
    Dec 2 '18 at 22:27














  • 1




    $begingroup$
    $F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
    $endgroup$
    – Will M.
    Dec 2 '18 at 22:27








1




1




$begingroup$
$F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
$endgroup$
– Will M.
Dec 2 '18 at 22:27




$begingroup$
$F_X(c + r) - F_X(c-r) = P(X in (c-r, c + r]) geq P(X = c),$ then let $r to 0.$ QED
$endgroup$
– Will M.
Dec 2 '18 at 22:27










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You have that $F_X(x) = Bbb P(X leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n downarrow x$ then $[X leq x] = bigcap_{n} [X leq x_n]$ implies that $$Bbb P(X leq x) = Bbb Pleft(bigcap_n [X leq x_n]right) = lim_{n to +infty} Bbb P(X leq x_n) = lim_{n to +infty} F_X(x_n).$$It follows that $F_X(c^+) doteq lim_{x to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $Bbb P(a< X leq b) = F_X(b)-F_X(a)$, and so $$Bbb P(X = c) = F_X(c) - lim_{x to c^-}F_X(x).$$Then $Bbb P(X = c)=0$ if and only if $F_X(c) = lim_{xto c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.






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    $begingroup$

    You have that $F_X(x) = Bbb P(X leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n downarrow x$ then $[X leq x] = bigcap_{n} [X leq x_n]$ implies that $$Bbb P(X leq x) = Bbb Pleft(bigcap_n [X leq x_n]right) = lim_{n to +infty} Bbb P(X leq x_n) = lim_{n to +infty} F_X(x_n).$$It follows that $F_X(c^+) doteq lim_{x to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $Bbb P(a< X leq b) = F_X(b)-F_X(a)$, and so $$Bbb P(X = c) = F_X(c) - lim_{x to c^-}F_X(x).$$Then $Bbb P(X = c)=0$ if and only if $F_X(c) = lim_{xto c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.






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      $begingroup$

      You have that $F_X(x) = Bbb P(X leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n downarrow x$ then $[X leq x] = bigcap_{n} [X leq x_n]$ implies that $$Bbb P(X leq x) = Bbb Pleft(bigcap_n [X leq x_n]right) = lim_{n to +infty} Bbb P(X leq x_n) = lim_{n to +infty} F_X(x_n).$$It follows that $F_X(c^+) doteq lim_{x to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $Bbb P(a< X leq b) = F_X(b)-F_X(a)$, and so $$Bbb P(X = c) = F_X(c) - lim_{x to c^-}F_X(x).$$Then $Bbb P(X = c)=0$ if and only if $F_X(c) = lim_{xto c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        You have that $F_X(x) = Bbb P(X leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n downarrow x$ then $[X leq x] = bigcap_{n} [X leq x_n]$ implies that $$Bbb P(X leq x) = Bbb Pleft(bigcap_n [X leq x_n]right) = lim_{n to +infty} Bbb P(X leq x_n) = lim_{n to +infty} F_X(x_n).$$It follows that $F_X(c^+) doteq lim_{x to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $Bbb P(a< X leq b) = F_X(b)-F_X(a)$, and so $$Bbb P(X = c) = F_X(c) - lim_{x to c^-}F_X(x).$$Then $Bbb P(X = c)=0$ if and only if $F_X(c) = lim_{xto c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.






        share|cite|improve this answer











        $endgroup$



        You have that $F_X(x) = Bbb P(X leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n downarrow x$ then $[X leq x] = bigcap_{n} [X leq x_n]$ implies that $$Bbb P(X leq x) = Bbb Pleft(bigcap_n [X leq x_n]right) = lim_{n to +infty} Bbb P(X leq x_n) = lim_{n to +infty} F_X(x_n).$$It follows that $F_X(c^+) doteq lim_{x to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $Bbb P(a< X leq b) = F_X(b)-F_X(a)$, and so $$Bbb P(X = c) = F_X(c) - lim_{x to c^-}F_X(x).$$Then $Bbb P(X = c)=0$ if and only if $F_X(c) = lim_{xto c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.







        share|cite|improve this answer














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        edited Dec 3 '18 at 1:45

























        answered Dec 2 '18 at 22:23









        Ivo TerekIvo Terek

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