I'm a bit confused with t vs x? So, when you start using t you said it's because it makes it less confusing but within the interval, it's still x as the variable, which makes me think they ARENT the same? Idk maybe im overthinking it
x and t are both values for the same random variable. I just didnt want to intergrate over x, and then plug x back in... some students get confused about plugging in x for x.... sorry if this confused you... though they are values for the same random variable.
You are thinking of the pdf which is the function f(x), which for t>2 is equal to 0. However, with the cdf, F(t), you're not measuring the value of the f(x), you're finding the integral of f(x) on the interval negative infinity to t. Because it includes all probability, the integral of f(x) on the interval negative infinity to t for t>2 must always equal 1.
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I'm a bit confused with t vs x? So, when you start using t you said it's because it makes it less confusing but within the interval, it's still x as the variable, which makes me think they ARENT the same? Idk maybe im overthinking it
x and t are both values for the same random variable. I just didnt want to intergrate over x, and then plug x back in... some students get confused about plugging in x for x.... sorry if this confused you... though they are values for the same random variable.
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How to find joint density function from cumulative distribution function
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Great explanation
but why t at 10:16 is less than 0. Why not t less than or equal to 0 (t
I think you can do it that way, but I don't think it matters if you use (
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Why was t>2 = to 1? Why not 0?
You are thinking of the pdf which is the function f(x), which for t>2 is equal to 0. However, with the cdf, F(t), you're not measuring the value of the f(x), you're finding the integral of f(x) on the interval negative infinity to t. Because it includes all probability, the integral of f(x) on the interval negative infinity to t for t>2 must always equal 1.
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