Mixing
Find the cumulative distribution function of Y = XII{X ≤ b}.
$begingroup$
Assume that X is a continuous and nonnegative random variable with the cumulative distribution function Fx. Let b > 0.
a) Find the cumulative distribution function of Y = XII{X ≤ b}.
b) Apply the general formula from (a) to exponential distribution with parameter
λ > 0.
I'm having trouble understanding the notation for Y = XII{X ≤ b} What does "II" mean?
probability probability-theory probability-distributions
asked Jan 17 at 17:26
Slam95Slam95
51
$endgroup$
add a comment |
$begingroup$
Assume that X is a continuous and nonnegative random variable with the cumulative distribution function Fx. Let b > 0.
a) Find the cumulative distribution function of Y = XII{X ≤ b}.
b) Apply the general formula from (a) to exponential distribution with parameter
λ > 0.
I'm having trouble understanding the notation for Y = XII{X ≤ b} What does "II" mean?
probability probability-theory probability-distributions
asked Jan 17 at 17:26
Slam95Slam95
51
$endgroup$
1
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02
add a comment |
$begingroup$
Assume that X is a continuous and nonnegative random variable with the cumulative distribution function Fx. Let b > 0.
a) Find the cumulative distribution function of Y = XII{X ≤ b}.
b) Apply the general formula from (a) to exponential distribution with parameter
λ > 0.
I'm having trouble understanding the notation for Y = XII{X ≤ b} What does "II" mean?
probability probability-theory probability-distributions
asked Jan 17 at 17:26
Slam95Slam95
51
$endgroup$
Assume that X is a continuous and nonnegative random variable with the cumulative distribution function Fx. Let b > 0.
a) Find the cumulative distribution function of Y = XII{X ≤ b}.
b) Apply the general formula from (a) to exponential distribution with parameter
λ > 0.
I'm having trouble understanding the notation for Y = XII{X ≤ b} What does "II" mean?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 17 at 17:26
Slam95Slam95
51
asked Jan 17 at 17:26
Slam95Slam95
51
asked Jan 17 at 17:26
Slam95Slam95
51
asked Jan 17 at 17:26
Slam95Slam95
51
asked Jan 17 at 17:26
Slam95Slam95
51
51
1
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02
add a comment |
1
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02
1
1
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
My guess is that
$$
Y=XI(Xleq b)=begin{cases}
X&text{if}, Xleq b\
0& text{if}, X>b
end{cases}
$$
so $Y$ equals $X$ truncated at $b$.
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
My guess is that
$$
Y=XI(Xleq b)=begin{cases}
X&text{if}, Xleq b\
0& text{if}, X>b
end{cases}
$$
so $Y$ equals $X$ truncated at $b$.
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
My guess is that
$$
Y=XI(Xleq b)=begin{cases}
X&text{if}, Xleq b\
0& text{if}, X>b
end{cases}
$$
so $Y$ equals $X$ truncated at $b$.
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
My guess is that
$$
Y=XI(Xleq b)=begin{cases}
X&text{if}, Xleq b\
0& text{if}, X>b
end{cases}
$$
so $Y$ equals $X$ truncated at $b$.
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
My guess is that
$$
Y=XI(Xleq b)=begin{cases}
X&text{if}, Xleq b\
0& text{if}, X>b
end{cases}
$$
so $Y$ equals $X$ truncated at $b$.
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 21:59
Foobaz JohnFoobaz John
22.2k41452
22.2k41452
add a comment |
add a comment |
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1
$begingroup$
X = 10. And XII = 12.
$endgroup$
– wolfies
Jan 17 at 17:46
$begingroup$
So it's like saying Y=X+2?
$endgroup$
– Slam95
Jan 17 at 18:02