Given an average return of 3.4% and a standard deviation of 7.4%, what's the
probability of losing money (ie, return less than zero)?

I'm interested in how to do the calculation as opposed the a specific
answer.

Thanks,
Fred

"Fred Smith" wrote:
Given an average return of 3.4% and a standard deviation
of 7.4%, what's the probability of losing money (ie,
return less than zero)?

You are asking "what is the probability that the cumulative distribution is
less than zero?".

The formulation depends on the distribution. Assuming a normal
distribution:

=NORMDIST(0, 3.4%, 7.4%, 1)

Note that for securities, the returns are assumed to be log normal. That
is, the distribution of the log of the returns is assumed to be normal, not
the distribution of the returns themselves. So 3.4% and 7.4% should be the
log mean and log std dev.

(It is debatable whether we should the arithmetic mean or the geometric mean
and corresponding std dev. Most experts tend to use the geometric mean,
which reflects the mean of the time series.)

But while the log normal assumption may be valid in general and over a long
period of time, it might not be valid for particular securities or data
series. Consequently, you might want look at the distribution of your data
and perhaps use BETADIST() instead.


----- original message -----

"Fred Smith" wrote in message
...
Given an average return of 3.4% and a standard deviation of 7.4%, what's
the probability of losing money (ie, return less than zero)?

I'm interested in how to do the calculation as opposed the a specific
answer.

Thanks,
Fred

Thanks, Joe. Just what I was looking for. Simple when you know how.

Fred.

"JoeU2004" wrote in message
...
"Fred Smith" wrote:
Given an average return of 3.4% and a standard deviation
of 7.4%, what's the probability of losing money (ie,
return less than zero)?

You are asking "what is the probability that the cumulative distribution
is less than zero?".

The formulation depends on the distribution. Assuming a normal
distribution:

=NORMDIST(0, 3.4%, 7.4%, 1)

Note that for securities, the returns are assumed to be log normal. That
is, the distribution of the log of the returns is assumed to be normal,
not the distribution of the returns themselves. So 3.4% and 7.4% should
be the log mean and log std dev.

(It is debatable whether we should the arithmetic mean or the geometric
mean and corresponding std dev. Most experts tend to use the geometric
mean, which reflects the mean of the time series.)

But while the log normal assumption may be valid in general and over a
long period of time, it might not be valid for particular securities or
data series. Consequently, you might want look at the distribution of
your data and perhaps use BETADIST() instead.


----- original message -----

"Fred Smith" wrote in message
...
Given an average return of 3.4% and a standard deviation of 7.4%, what's
the probability of losing money (ie, return less than zero)?

I'm interested in how to do the calculation as opposed the a specific
answer.

Thanks,
Fred

"Fred Smith" wrote:
Thanks, Joe. Just what I was looking for.

You're welcome. Thanks for the feedback.


Simple when you know how.

Arguably, deceptively so.

If you figure out how to use BETADIST for a particular distribution, let
me know. Send email to joeu2004 "at" hotmail.com.


I wrote:
It is debatable whether we should the arithmetic mean or the geometric
mean and corresponding std dev. Most experts tend to use the geometric
mean, which reflects the mean of the time series.

Of course, we use the geometic mean to compute the average return of a time
series.

However, I think we should use the arithmetic mean and std dev of the log
returns, since then we are looking at the distribution of returns as
statistics, not the time series.

But I frequently see the geometric std dev of the time series of log
returns, even of the returns themselves, used to define "volatility". And
that volatility factor, along with the geometric mean of the time series (of
the log returns or of the returns) are used in Monte Carlo simulations as
well as other statistical analysis.

In fact, as I recall vaguely (it's been a few years), the leading texts on
the mean-variance theory use the geometric mean and std dev.

And I was told that the PhDs responsible for the Fidelity Retirement Income
Planner were recently (well, within the past 18 months now) convinced to
switch from the arithmetic mean and std dev to the geometric mean and std
dev. (I'm told that they were convinced by former colleagues of mine, who
are computer engineers, not financial engineers.)

I wonder what Mike Middleton has to say about all this.


----- original message -----

"Fred Smith" wrote in message
...
Thanks, Joe. Just what I was looking for. Simple when you know how.

Fred.

"JoeU2004" wrote in message
...
"Fred Smith" wrote:
Given an average return of 3.4% and a standard deviation
of 7.4%, what's the probability of losing money (ie,
return less than zero)?

You are asking "what is the probability that the cumulative distribution
is less than zero?".

The formulation depends on the distribution. Assuming a normal
distribution:

=NORMDIST(0, 3.4%, 7.4%, 1)

Note that for securities, the returns are assumed to be log normal. That
is, the distribution of the log of the returns is assumed to be normal,
not the distribution of the returns themselves. So 3.4% and 7.4% should
be the log mean and log std dev.

(It is debatable whether we should the arithmetic mean or the geometric
mean and corresponding std dev. Most experts tend to use the geometric
mean, which reflects the mean of the time series.)

But while the log normal assumption may be valid in general and over a
long period of time, it might not be valid for particular securities or
data series. Consequently, you might want look at the distribution of
your data and perhaps use BETADIST() instead.


----- original message -----

"Fred Smith" wrote in message
...
Given an average return of 3.4% and a standard deviation of 7.4%, what's
the probability of losing money (ie, return less than zero)?

I'm interested in how to do the calculation as opposed the a specific
answer.

Thanks,
Fred

Errata (aarrgghh!)....

I wrote (I wish I could say I was stoned at the time):
But I frequently see the geometric std dev of the time
series of log returns, even of the returns themselves,
used to define "volatility". And that volatility factor,
along with the geometric mean of the time series (of the
log returns or of the returns) are used in Monte Carlo
simulations as well as other statistical analysis.

The issue is simply whether to use the arithmetic mean and std dev of the
log returns or the geometric mean and std dev of the returns in a normal
distribution. (Some people even use the arithmetic mean and std dev of the
returns.)

Of course, the antilogs of the arithmetic mean and std dev of the log
returns are the geometric mean and std dev of the returns. No one uses the
geometric mean and std dev of the "time series of log returns". That's
nonsense!


----- original message -----

"JoeU2004" wrote in message
...
"Fred Smith" wrote:
Thanks, Joe. Just what I was looking for.

You're welcome. Thanks for the feedback.


Simple when you know how.

Arguably, deceptively so.

If you figure out how to use BETADIST for a particular distribution, let
me know. Send email to joeu2004 "at" hotmail.com.


I wrote:
It is debatable whether we should the arithmetic mean or the geometric
mean and corresponding std dev. Most experts tend to use the geometric
mean, which reflects the mean of the time series.

Of course, we use the geometic mean to compute the average return of a
time series.

However, I think we should use the arithmetic mean and std dev of the log
returns, since then we are looking at the distribution of returns as
statistics, not the time series.

But I frequently see the geometric std dev of the time series of log
returns, even of the returns themselves, used to define "volatility". And
that volatility factor, along with the geometric mean of the time series
(of the log returns or of the returns) are used in Monte Carlo simulations
as well as other statistical analysis.

In fact, as I recall vaguely (it's been a few years), the leading texts on
the mean-variance theory use the geometric mean and std dev.

And I was told that the PhDs responsible for the Fidelity Retirement
Income Planner were recently (well, within the past 18 months now)
convinced to switch from the arithmetic mean and std dev to the geometric
mean and std dev. (I'm told that they were convinced by former colleagues
of mine, who are computer engineers, not financial engineers.)

I wonder what Mike Middleton has to say about all this.


----- original message -----

"Fred Smith" wrote in message
...
Thanks, Joe. Just what I was looking for. Simple when you know how.

Fred.

"JoeU2004" wrote in message
...
"Fred Smith" wrote:
Given an average return of 3.4% and a standard deviation
of 7.4%, what's the probability of losing money (ie,
return less than zero)?

You are asking "what is the probability that the cumulative distribution
is less than zero?".

The formulation depends on the distribution. Assuming a normal
distribution:

=NORMDIST(0, 3.4%, 7.4%, 1)

Note that for securities, the returns are assumed to be log normal.
That is, the distribution of the log of the returns is assumed to be
normal, not the distribution of the returns themselves. So 3.4% and
7.4% should be the log mean and log std dev.

(It is debatable whether we should the arithmetic mean or the geometric
mean and corresponding std dev. Most experts tend to use the geometric
mean, which reflects the mean of the time series.)

But while the log normal assumption may be valid in general and over a
long period of time, it might not be valid for particular securities or
data series. Consequently, you might want look at the distribution of
your data and perhaps use BETADIST() instead.


----- original message -----

"Fred Smith" wrote in message
...
Given an average return of 3.4% and a standard deviation of 7.4%,
what's the probability of losing money (ie, return less than zero)?

I'm interested in how to do the calculation as opposed the a specific
answer.

Thanks,
Fred