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Ben
Ben
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I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$. We may not assume independence.

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$. We may not assume independence.

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.

I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$. We may not assume independence.

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.

I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$. We may not assume independence.

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

updated question to make it clear that you are to prove or disprove and added that independence may not be assumed.
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StatCurious
StatCurious
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I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

IfProve or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?. We may not assume independence.

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.

I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.

I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$. We may not assume independence.

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.

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StatCurious
StatCurious
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If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?

I came across this question in a review of an old exam I took. I didn't get the answer correctly then, and I'm struggling to figure the answer out now. Can anyone help me reason through this?

If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?

Here is what I attempted:

I figured I might be able to approach this by proving this through contradiction. I started by assuming $P(X<Y)=0$. Then,

\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\ & = & 0+P(X\le z,X\ge Y) \end{eqnarray*}

Can anyone help from here?

Thanks.