How to prove the following probability question?

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Suppose we have $widehatM=(widehatr_a,widehatr_b,widehatB)$ and $M_0=(r_a,r_b,B)$. Show that $P(widehatM=M_0)to 1$ if and only if $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$.



I think it is obvious because $P(widehatM=M_0)to 1$ can be written as $P(widehatr_a=r_a,widehatr_b=r_b,widehatB=B)to 1$, then we have $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$. But my professor said it is wrong. Can you please help me with it?




probability




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  • Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
    – Robert Howard
    Jul 18 at 21:41










  • What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
    – Arnaud Mortier
    Jul 18 at 21:57










  • The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
    – Jeff
    Jul 18 at 22:01















up vote
-2
down vote

favorite












Suppose we have $widehatM=(widehatr_a,widehatr_b,widehatB)$ and $M_0=(r_a,r_b,B)$. Show that $P(widehatM=M_0)to 1$ if and only if $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$.



I think it is obvious because $P(widehatM=M_0)to 1$ can be written as $P(widehatr_a=r_a,widehatr_b=r_b,widehatB=B)to 1$, then we have $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$. But my professor said it is wrong. Can you please help me with it?




probability




share|cite|improve this question





















  • Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
    – Robert Howard
    Jul 18 at 21:41










  • What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
    – Arnaud Mortier
    Jul 18 at 21:57










  • The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
    – Jeff
    Jul 18 at 22:01













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











Suppose we have $widehatM=(widehatr_a,widehatr_b,widehatB)$ and $M_0=(r_a,r_b,B)$. Show that $P(widehatM=M_0)to 1$ if and only if $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$.



I think it is obvious because $P(widehatM=M_0)to 1$ can be written as $P(widehatr_a=r_a,widehatr_b=r_b,widehatB=B)to 1$, then we have $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$. But my professor said it is wrong. Can you please help me with it?




probability




share|cite|improve this question













Suppose we have $widehatM=(widehatr_a,widehatr_b,widehatB)$ and $M_0=(r_a,r_b,B)$. Show that $P(widehatM=M_0)to 1$ if and only if $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$.



I think it is obvious because $P(widehatM=M_0)to 1$ can be written as $P(widehatr_a=r_a,widehatr_b=r_b,widehatB=B)to 1$, then we have $P(widehatr_a=r_a)to 1$, $P(widehatr_b=r_b)to 1$, and $P(widehatB=B)to 1$. But my professor said it is wrong. Can you please help me with it?





probability





share|cite|improve this question












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edited Jul 18 at 21:45
























asked Jul 18 at 21:39









Jeff

11




11











  • Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
    – Robert Howard
    Jul 18 at 21:41










  • What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
    – Arnaud Mortier
    Jul 18 at 21:57










  • The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
    – Jeff
    Jul 18 at 22:01

















  • Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
    – Robert Howard
    Jul 18 at 21:41










  • What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
    – Arnaud Mortier
    Jul 18 at 21:57










  • The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
    – Jeff
    Jul 18 at 22:01
















Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
– Robert Howard
Jul 18 at 21:41




Welcome to MSE! What do you think might be a good way to start on this problem? We'll be happy to help you along if you show us that you've made an effort, but we won't blindly do your work for you.
– Robert Howard
Jul 18 at 21:41












What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
– Arnaud Mortier
Jul 18 at 21:57




What do all these letters stand for? Convergence is in what sense? How is the distribution defined that allows you to talk about probabilities?
– Arnaud Mortier
Jul 18 at 21:57












The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
– Jeff
Jul 18 at 22:01





The letter M with the hat is the estimator of M_0. Convergence in probability to 1 in all cases.
– Jeff
Jul 18 at 22:01
















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