Calculating Standard Deviation for a Probability Density Function?

[Schwarzschild90]

Homework Statement

Homework Equations

See below

The Attempt at a Solution

\begin{align}
\begin{split}
p(x) = C \ x \ exp(-x/ \lambda)
\end{split}
\end{align}

If $p(x)$ is a probability density function on the interval $ 0 \textless x \textless + \infty $ , then it follows that the normalization constant can be isolated by setting the area under the curve equal to one

\begin{align}
\begin{split}
\int_0^\infty p(x) \ dx = 1 \to \\ C \int_0^\infty x \ exp(-x/ \lambda) \ dx = 1 \to \\ C \lambda^2 = 1 \to \\ C = \frac{1}{\lambda^2}
\end{split}
\end{align}

\subsection*{(b)}
The mean $\mu$ (or expection $E(X)$) of X is

\begin{align}
\begin{split}
mean = E(X) = \int_a^b x \ f(x) \ dx = \frac{1}{\lambda^2} \int_0^\infty x^2 \ exp(-x/ \lambda) \ dx = 2 \lambda^3
\end{split}
\end{align}

\subsection*{(c)}
Now, determine the standard deviation $\sigma$ of X

\begin{align}
\begin{split}
\sigma =
\end{split}
\end{align}

 

[haruspex]

In 3), did you forget the factor C?
Are you stuck on the standard deviation, or did your post just get truncated?
What formulae do you know related to variance and s.d.?

 

[Schwarzschild90]

I imagined that someone could see if what I had already done was correct

I know the following formula for variance: (E(X^2) - mu^2)

 

[Ray Vickson]

Homework Statement

View attachment 95542

Homework Equations

See below

The Attempt at a Solution

\begin{align}
\begin{split}
p(x) = C \ x \ exp(-x/ \lambda)
\end{split}
\end{align}

If $p(x)$ is a probability density function on the interval $ 0 \textless x \textless + \infty $ , then it follows that the normalization constant can be isolated by setting the area under the curve equal to one

\begin{align}
\begin{split}
\int_0^\infty p(x) \ dx = 1 \to \\ C \int_0^\infty x \ exp(-x/ \lambda) \ dx = 1 \to \\ C \lambda^2 = 1 \to \\ C = \frac{1}{\lambda^2}
\end{split}
\end{align}

\subsection*{(b)}
The mean $\mu$ (or expection $E(X)$) of X is

\begin{align}
\begin{split}
mean = E(X) = \int_a^b x \ f(x) \ dx = \frac{1}{\lambda^2} \int_0^\infty x^2 \ exp(-x/ \lambda) \ dx = 2 \lambda^3
\end{split}
\end{align}

\subsection*{(c)}
Now, determine the standard deviation $\sigma$ of X

\begin{align}
\begin{split}
\sigma =
\end{split}
\end{align}

Your computation of $EX$ is incorrect.

Hint: check dimensions. If $X$ happened to be a time, $\lambda$ would have dimensions of time also, and so your $\lambda^3$ would have dimensions of $\text{time}^3$ (but should be just 'time').

 

[Schwarzschild90]

I forgot to divide by λ².

Now, does 2/λ seem reasonable in your eyes?

 

[Ray Vickson]

I forgot to divide by λ².

Now, does 2/λ seem reasonable in your eyes?

Both reasonable and correct, not only in my eyes but absolutely.

 

[haruspex]

I forgot to divide by λ².

Now, does 2/λ seem reasonable in your eyes?

Now you seem to have divided by λ⁴.
If you replace x with λu in the integral you can see you should get a factor λ and an integral independent of λ.

I
I know the following formula for variance: (E(X^2) - mu^2)

So turn that into an integral. You already have μ.

 

[Ray Vickson]

Both reasonable and correct, not only in my eyes but absolutely.

Sorry: I am so used to looking at such distributions in the form $c t e^{-\lambda t}$ (that is, with $\lambda t$ instead of $t/\lambda$) that my brain failed to process the difference in your case. So, NO: you are still not correct.

 

[Schwarzschild90]

Like this? I don't understand why we replace x with lambda u

\begin{align}
\begin{split}
mean = E(X) = \int_a^b x \ f(x) \ dx = \frac{1}{\lambda^2} \int_0^\infty (\lambda u)^2 \ exp(-(\lambda u)/ \lambda) \ dx =
\end{split}
\end{align}

 

[haruspex]

Like this? I don't understand why we replace x with lambda u

\begin{align}
\begin{split}
mean = E(X) = \int_a^b x \ f(x) \ dx = \frac{1}{\lambda^2} \int_0^\infty (\lambda u)^2 \ exp(-(\lambda u)/ \lambda) \ dx =
\end{split}
\end{align}

I was just using that to show that the final expression must be linear in lambda.
$\frac{1}{\lambda^2} \int_0^\infty (\lambda u)^2 \ exp(-(\lambda u)/ \lambda) \ d(\lambda u)=\lambda\int_0^\infty u^2e^{-u}.du$. Your first version ended up with a factor λ³ because you forgot the factor C=λ^-2, but then somehow you got λ^-1, as though you made the correction twice over.

 

[Schwarzschild90]

So, 2\lambda would be correct?

 

[haruspex]

So, 2\lambda would be correct?

Yes.

 

[Schwarzschild90]

Will use this formula to calculate the standard deviation
\begin{align}
\begin{split}
\sigma = std(X) =\sqrt{ \int_a^b (x-\mu)^2 \ f(x) \ dx }
\end{split}
\end{align}

 

[haruspex]

Will use this formula to calculate the standard deviation
\begin{align}
\begin{split}
\sigma = std(X) =\sqrt{ \int_a^b (x-\mu)^2 \ f(x) \ dx }
\end{split}
\end{align}

Ok, but you might find the form var(X)=E(X²)-E(X)² more convenient than E((X-E(X))²).

 

[Schwarzschild90]

I assume sqrt(E(X2)-E(X)2)?

 

[haruspex]

I assume sqrt(E(X2)-E(X)2)?

For standard deviation, yes. I was quoting formulae for variance (σ²).

 

[Schwarzschild90]

\begin{align}
\begin{split}
E(X^2) = \frac{1}{\lambda^2} \int_0^\infty x^3 \ exp(-x/ \lambda) \ dx = 6 \lambda^2 \\
\mu^2 = 4\lambda^2 \\
\sigma = std(X) = \sqrt{var(X)} = \sqrt{E(X^2) - \mu^2} = \sqrt{6\lambda^2 - 4\lambda^2}= \sqrt{2\lambda^2} = \sqrt{2}\lambda \\
\end{split}
\end{align}

It seems almost impossible to commit a mistake using the above equation :P, but I've been wrong before

 

[haruspex]

\begin{align}
\begin{split}
E(X^2) = \frac{1}{\lambda^2} \int_0^\infty x^3 \ exp(-x/ \lambda) \ dx = 6 \lambda^2 \\
\mu^2 = 4\lambda^2 \\
\sigma = std(X) = \sqrt{var(X)} = \sqrt{E(X^2) - \mu^2} = \sqrt{6\lambda^2 - 4\lambda^2}= \sqrt{2\lambda^2} = \sqrt{2}\lambda \\
\end{split}
\end{align}

It seems almost impossible to commit a mistake using the above equation :P, but I've been wrong before

Looks right.