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Stuck on both.
2b shpuld I include the x equalling infinity? Its not really possible but didnt know what to write.

What do I do on 3? I just took a guess with my working


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Original post by cooldudeman
Stuck on both.
2b shpuld I include the x equalling infinity? Its not really possible but didnt know what to write.


It's a cdf, so lim as x goes to infinity is always 1.

You have a bigger problem. The cdf should never go down as x increases. Can you see what you've missed out?

For 3.

It would help to draw a diagram, with X1 and X2 being the axes. You want max(X1,X2} = Y,

Let Y equal any value and try plotting the valid region.
(edited 10 years ago)
Reply 2
Original post by ghostwalker
It's a cdf, so lim as x goes to infinity is always 1.

You have a bigger problem. The cdf should never go down as x increases. Can you see what you've missed out?

For 3.

It would help to draw a diagram, with X1 and X2 being the axes. You want max(X1,X2} = Y,

Let Y equal any value and try plotting the valid region.


For 2, I cant see. Was I.meant to integrate the -infin <x<0 part with limits x to 0? Instead of minus inf to x like I did.

For 3, I dont understand. I drew a diagram loke.you said. With (1,1) being the top right corner. Like a box of area 1. The pdf is just Y~U(0,1) isnt it. And for tje P (Y<=y) part, im lost.

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Original post by cooldudeman
For 2, I cant see. Was I.meant to integrate the -infin <x<0 part with limits x to 0? Instead of minus inf to x like I did.


No.

For parts x>0 you need to add in the intergral from -infinty to 0 to what you're already doing.

F(x) = integral from minus infinity to x.


For 3, I dont understand. I drew a diagram loke.you said. With (1,1) being the top right corner. Like a box of area 1. The pdf is just Y~U(0,1) isnt it. And for tje P (Y<=y) part, im lost.


Set Y to a specific value, say 1/2. Then what area in your diagram represent Y<1/2?

If Y < 1/2, then what are the possibilities of X1 and X2?
Reply 4
Original post by ghostwalker
No.

For parts x>0 you need to add in the intergral from -infinty to 0 to what you're already doing.

F(x) = integral from minus infinity to x.



Set Y to a specific value, say 1/2. Then what area in your diagram represent Y<1/2?

If Y < 1/2, then what are the possibilities of X1 and X2?


Really confused with 2. You mean like this?

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Original post by cooldudeman
Really confused with 2. You mean like this?

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Ish.

F(s)=sf(x)  dx\displaystyle F(s) = \int_{-\infty}^s f(x)\;dx

But because f(x) is defined in two pieces; if s is greater than zero, then, this works out to be:

F(s)=0cex  dx+0scex  dx\displaystyle F(s) = \int_{-\infty}^0 ce^x\;dx+\int^{s}_0ce^{-x}\;dx
Reply 6
Original post by ghostwalker
Ish.

F(s)=sf(x)  dx\displaystyle F(s) = \int_{-\infty}^s f(x)\;dx

But because f(x) is defined in two pieces; if s is greater than zero, then, this works out to be:

F(s)=0cex  dx+0scex  dx\displaystyle F(s) = \int_{-\infty}^0 ce^x\;dx+\int^{s}_0ce^{-x}\;dx


Can ypu explain how you figured that out? So I know that the cdf should always be increasing with x increasing.
And this pdf graph looked like a symmetrical exp graph. How did you know to chamge the limits like that

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Original post by cooldudeman
Can ypu explain how you figured that out? So I know that the cdf should always be increasing with x increasing.
And this pdf graph looked like a symmetrical exp graph. How did you know to chamge the limits like that

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The cdf can never be decreasing as the pdf (f(x) can never be negative. The cdf can be constant in part.

This is just the integral of f(x) from -infinity to s. It's split into two because f(x) is defined differently on the two intevals.

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