how to justify the probability of a linear transformation of a event?which rule can be used to compute the probability of $P a<Xleq b $?Derive the marginal probability function for XDeriving the formula for transforming random variablesProbability of an event to occurAbout the difference between “probability” and “probability mass function”How to calculate the probability of this normally distributed event?Probability of an event that has happened, to have happened in a specific time range?Find PMF, Expectation and Variance.Probability problem(pmf, avg value, var, allocation) : hitting an enemy until he diesBishop - Pattern Recognition & Machine Learning, Exercise 1.4Probability distribution vs. probability mass function / Probability density function terms: what's the difference
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how to justify the probability of a linear transformation of a event?
which rule can be used to compute the probability of $P a<Xleq b $?Derive the marginal probability function for XDeriving the formula for transforming random variablesProbability of an event to occurAbout the difference between “probability” and “probability mass function”How to calculate the probability of this normally distributed event?Probability of an event that has happened, to have happened in a specific time range?Find PMF, Expectation and Variance.Probability problem(pmf, avg value, var, allocation) : hitting an enemy until he diesBishop - Pattern Recognition & Machine Learning, Exercise 1.4Probability distribution vs. probability mass function / Probability density function terms: what's the difference
$begingroup$
I am confused when I trying to understand this discussion
I have already known
the probability of a linear function $Y = g(X) = aX + b$, where X is a random variable with a PMF P_X(x)
is
$$
P_Y(y) =
sumlimits_x P_X(x)
$$
how to justify the probability equation of a event like this?
$$
P(
X > a
) =
P(
dfracX-musigma >
dfraca-musigma)
)
$$
probability
asked Mar 22 at 8:57
brennnbrennn
132
$endgroup$
add a comment |
$begingroup$
I am confused when I trying to understand this discussion
I have already known
the probability of a linear function $Y = g(X) = aX + b$, where X is a random variable with a PMF P_X(x)
is
$$
P_Y(y) =
sumlimits_x P_X(x)
$$
how to justify the probability equation of a event like this?
$$
P(
X > a
) =
P(
dfracX-musigma >
dfraca-musigma)
)
$$
probability
asked Mar 22 at 8:57
brennnbrennn
132
$endgroup$
add a comment |
$begingroup$
I am confused when I trying to understand this discussion
I have already known
the probability of a linear function $Y = g(X) = aX + b$, where X is a random variable with a PMF P_X(x)
is
$$
P_Y(y) =
sumlimits_x P_X(x)
$$
how to justify the probability equation of a event like this?
$$
P(
X > a
) =
P(
dfracX-musigma >
dfraca-musigma)
)
$$
probability
asked Mar 22 at 8:57
brennnbrennn
132
$endgroup$
I am confused when I trying to understand this discussion
I have already known
the probability of a linear function $Y = g(X) = aX + b$, where X is a random variable with a PMF P_X(x)
is
$$
P_Y(y) =
sumlimits_x P_X(x)
$$
how to justify the probability equation of a event like this?
$$
P(
X > a
) =
P(
dfracX-musigma >
dfraca-musigma)
)
$$
probability
probability
asked Mar 22 at 8:57
brennnbrennn
132
asked Mar 22 at 8:57
brennnbrennn
132
asked Mar 22 at 8:57
brennnbrennn
132
asked Mar 22 at 8:57
brennnbrennn
132
asked Mar 22 at 8:57
brennnbrennn
132
132
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is because of simple algebra (for $sigma > 0$): $$beginalign
x>a &iff x-mu > a-mu \
&iff fracx-musigma > fraca-musigma.
endalign$$
So $X> a$ occurs if and only if $leftfracX-musigma > fraca-musigmaright$ occurs. Thus the two probabilities are the same.
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
$endgroup$
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
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
$begingroup$
It is because of simple algebra (for $sigma > 0$): $$beginalign
x>a &iff x-mu > a-mu \
&iff fracx-musigma > fraca-musigma.
endalign$$
So $X> a$ occurs if and only if $leftfracX-musigma > fraca-musigmaright$ occurs. Thus the two probabilities are the same.
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
$endgroup$
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
add a comment |
$begingroup$
It is because of simple algebra (for $sigma > 0$): $$beginalign
x>a &iff x-mu > a-mu \
&iff fracx-musigma > fraca-musigma.
endalign$$
So $X> a$ occurs if and only if $leftfracX-musigma > fraca-musigmaright$ occurs. Thus the two probabilities are the same.
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
$endgroup$
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
add a comment |
$begingroup$
It is because of simple algebra (for $sigma > 0$): $$beginalign
x>a &iff x-mu > a-mu \
&iff fracx-musigma > fraca-musigma.
endalign$$
So $X> a$ occurs if and only if $leftfracX-musigma > fraca-musigmaright$ occurs. Thus the two probabilities are the same.
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
$endgroup$
It is because of simple algebra (for $sigma > 0$): $$beginalign
x>a &iff x-mu > a-mu \
&iff fracx-musigma > fraca-musigma.
endalign$$
So $X> a$ occurs if and only if $leftfracX-musigma > fraca-musigmaright$ occurs. Thus the two probabilities are the same.
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
answered Mar 22 at 9:02
Minus One-TwelfthMinus One-Twelfth
3,328413
3,328413
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
add a comment |
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
$begingroup$
thanks, I helps. and would you please give a real life example to illustrate this transformation? i am having some difficulties to get a complete understanding of it.
$endgroup$
– brennn
Mar 22 at 9:06
1
1
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
Maybe just consider some numbers. Imagine for now that $mu=0$. So you need to understand $x>aiff x/sigma > a/sigma$. Could you see why this would be true if $sigma =2$ for instance (so $x>a iff x/2 > a/2$)?
$endgroup$
– Minus One-Twelfth
Mar 22 at 9:09
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
$begingroup$
What is so difficult? All outcomes where $X>a$ are outcomes where $(X-mu)/sigma>(a-mu)/sigma$ because of the algebra: $x>aiff (x-b)/c>(a-b)/c$ (when $c>0$) $$omega:X(omega)>a=omega:(X(omega)-mu)/sigma=(a-mu)/sigma$$
$endgroup$
– Graham Kemp
Mar 22 at 9:13
add a comment |
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