$min{X, Y}$ is geometrically distributed according to parameter $1 - (1-p)^{2}$












0














Let $X,Y$ be independent from the parameter $0 < p < 1$ distributed random variable on a discrete probability space $(Omega, mathcal{F}, mathbb P)$. Show that $min{X,Y}$ is geometrically distrbuted to $1-(1-p)^{2}$.



Steps:
I am having trouble understanding what it is my end result has to be, in order to show that it is indeed geometrically distributed. I mean for a geometric distribution $mathbb P(k) = p(1-p)^{k-1}$ for parameter $p$, but how exactly am I showing a geometric distribution?



$mathbb P( min{X,Y} > k)=mathbb P((X>k) cap (Y>k))$ and then what?



In terms of the parameter, I've got: $(1-(1-p)^{2})(1-(1-(1-p)^{2}))=p^{2}(2-p)^{2}$



I do not understand how the two work together...






real-analysis probability-theory stochastic-processes geometric-probability







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    0














    Let $X,Y$ be independent from the parameter $0 < p < 1$ distributed random variable on a discrete probability space $(Omega, mathcal{F}, mathbb P)$. Show that $min{X,Y}$ is geometrically distrbuted to $1-(1-p)^{2}$.



    Steps:
    I am having trouble understanding what it is my end result has to be, in order to show that it is indeed geometrically distributed. I mean for a geometric distribution $mathbb P(k) = p(1-p)^{k-1}$ for parameter $p$, but how exactly am I showing a geometric distribution?



    $mathbb P( min{X,Y} > k)=mathbb P((X>k) cap (Y>k))$ and then what?



    In terms of the parameter, I've got: $(1-(1-p)^{2})(1-(1-(1-p)^{2}))=p^{2}(2-p)^{2}$



    I do not understand how the two work together...






    real-analysis probability-theory stochastic-processes geometric-probability







    share|cite|improve this question

























      0












      0








      0


      1





      Let $X,Y$ be independent from the parameter $0 < p < 1$ distributed random variable on a discrete probability space $(Omega, mathcal{F}, mathbb P)$. Show that $min{X,Y}$ is geometrically distrbuted to $1-(1-p)^{2}$.



      Steps:
      I am having trouble understanding what it is my end result has to be, in order to show that it is indeed geometrically distributed. I mean for a geometric distribution $mathbb P(k) = p(1-p)^{k-1}$ for parameter $p$, but how exactly am I showing a geometric distribution?



      $mathbb P( min{X,Y} > k)=mathbb P((X>k) cap (Y>k))$ and then what?



      In terms of the parameter, I've got: $(1-(1-p)^{2})(1-(1-(1-p)^{2}))=p^{2}(2-p)^{2}$



      I do not understand how the two work together...






      real-analysis probability-theory stochastic-processes geometric-probability







      share|cite|improve this question













      Let $X,Y$ be independent from the parameter $0 < p < 1$ distributed random variable on a discrete probability space $(Omega, mathcal{F}, mathbb P)$. Show that $min{X,Y}$ is geometrically distrbuted to $1-(1-p)^{2}$.



      Steps:
      I am having trouble understanding what it is my end result has to be, in order to show that it is indeed geometrically distributed. I mean for a geometric distribution $mathbb P(k) = p(1-p)^{k-1}$ for parameter $p$, but how exactly am I showing a geometric distribution?



      $mathbb P( min{X,Y} > k)=mathbb P((X>k) cap (Y>k))$ and then what?



      In terms of the parameter, I've got: $(1-(1-p)^{2})(1-(1-(1-p)^{2}))=p^{2}(2-p)^{2}$



      I do not understand how the two work together...






      real-analysis probability-theory stochastic-processes geometric-probability




      real-analysis probability-theory stochastic-processes geometric-probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 18 at 14:09









      SABOY

      549311




      549311






















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          begin{align}
          P(min(X,Y) > k) &= P(X > k, Y>k) \
          &= P(X > k)P(Y>k) \
          &= P(X >k)^2 \
          &=((1-p)^2)^{k}
          end{align}
          Hence the CDF is $1-((1-p)^2)^k=1-(1-color{blue}{(1-(1-p)^2)})^k$



          Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$






          share|cite|improve this answer





















          • Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
            – SABOY
            Nov 18 at 14:26












          • $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
            – Siong Thye Goh
            Nov 18 at 14:29










          • In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
            – SABOY
            Nov 18 at 14:31






          • 1




            they are both geometric distribution with probability $p$ isn't it?
            – Siong Thye Goh
            Nov 18 at 14:33










          • But the OP didn't actually state that.
            – Robert Israel
            Nov 18 at 14:35











          Your Answer





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          1 Answer
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          active

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          3














          begin{align}
          P(min(X,Y) > k) &= P(X > k, Y>k) \
          &= P(X > k)P(Y>k) \
          &= P(X >k)^2 \
          &=((1-p)^2)^{k}
          end{align}
          Hence the CDF is $1-((1-p)^2)^k=1-(1-color{blue}{(1-(1-p)^2)})^k$



          Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$






          share|cite|improve this answer





















          • Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
            – SABOY
            Nov 18 at 14:26












          • $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
            – Siong Thye Goh
            Nov 18 at 14:29










          • In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
            – SABOY
            Nov 18 at 14:31






          • 1




            they are both geometric distribution with probability $p$ isn't it?
            – Siong Thye Goh
            Nov 18 at 14:33










          • But the OP didn't actually state that.
            – Robert Israel
            Nov 18 at 14:35
















          3














          begin{align}
          P(min(X,Y) > k) &= P(X > k, Y>k) \
          &= P(X > k)P(Y>k) \
          &= P(X >k)^2 \
          &=((1-p)^2)^{k}
          end{align}
          Hence the CDF is $1-((1-p)^2)^k=1-(1-color{blue}{(1-(1-p)^2)})^k$



          Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$






          share|cite|improve this answer





















          • Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
            – SABOY
            Nov 18 at 14:26












          • $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
            – Siong Thye Goh
            Nov 18 at 14:29










          • In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
            – SABOY
            Nov 18 at 14:31






          • 1




            they are both geometric distribution with probability $p$ isn't it?
            – Siong Thye Goh
            Nov 18 at 14:33










          • But the OP didn't actually state that.
            – Robert Israel
            Nov 18 at 14:35














          3












          3








          3






          begin{align}
          P(min(X,Y) > k) &= P(X > k, Y>k) \
          &= P(X > k)P(Y>k) \
          &= P(X >k)^2 \
          &=((1-p)^2)^{k}
          end{align}
          Hence the CDF is $1-((1-p)^2)^k=1-(1-color{blue}{(1-(1-p)^2)})^k$



          Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$






          share|cite|improve this answer












          begin{align}
          P(min(X,Y) > k) &= P(X > k, Y>k) \
          &= P(X > k)P(Y>k) \
          &= P(X >k)^2 \
          &=((1-p)^2)^{k}
          end{align}
          Hence the CDF is $1-((1-p)^2)^k=1-(1-color{blue}{(1-(1-p)^2)})^k$



          Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 14:17









          Siong Thye Goh

          99.1k1464117




          99.1k1464117












          • Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
            – SABOY
            Nov 18 at 14:26












          • $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
            – Siong Thye Goh
            Nov 18 at 14:29










          • In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
            – SABOY
            Nov 18 at 14:31






          • 1




            they are both geometric distribution with probability $p$ isn't it?
            – Siong Thye Goh
            Nov 18 at 14:33










          • But the OP didn't actually state that.
            – Robert Israel
            Nov 18 at 14:35


















          • Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
            – SABOY
            Nov 18 at 14:26












          • $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
            – Siong Thye Goh
            Nov 18 at 14:29










          • In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
            – SABOY
            Nov 18 at 14:31






          • 1




            they are both geometric distribution with probability $p$ isn't it?
            – Siong Thye Goh
            Nov 18 at 14:33










          • But the OP didn't actually state that.
            – Robert Israel
            Nov 18 at 14:35
















          Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
          – SABOY
          Nov 18 at 14:26






          Thank you! Please just remind me why we can assume $P(X > k)P(Y > k)= P(X > k)^{2}$. And then adding on to that why would it be equal to $((1-p)^{2})^{k}$?
          – SABOY
          Nov 18 at 14:26














          $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
          – Siong Thye Goh
          Nov 18 at 14:29




          $X$ and $Y$ are IID. Also we have $P(X > k)=(1-p)^k$.
          – Siong Thye Goh
          Nov 18 at 14:29












          In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
          – SABOY
          Nov 18 at 14:31




          In the pre-text, it was only stated that the $X, Y$ are independent. Can I automatically assume IID?
          – SABOY
          Nov 18 at 14:31




          1




          1




          they are both geometric distribution with probability $p$ isn't it?
          – Siong Thye Goh
          Nov 18 at 14:33




          they are both geometric distribution with probability $p$ isn't it?
          – Siong Thye Goh
          Nov 18 at 14:33












          But the OP didn't actually state that.
          – Robert Israel
          Nov 18 at 14:35




          But the OP didn't actually state that.
          – Robert Israel
          Nov 18 at 14:35


















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