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Discrete-continuous random variable!

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone!

I'm looking at the following random variables:

$Z_1$ is normally distributed with zero mean and variance $\sigma _1 ^2$
$Z_2$ is normally distributed with zero mean and variance $\sigma _2 ^2$

$B$ is Bernoulli with probability of success $p$.

$X$ is a random variable that takes $Z_1$ if $B=1$ and $Z_2$ if $B=0$.

What is the variance of $X$?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: Dicrete-continuous random variable!

Hello everyone!

I'm looking at the following random variables:

$Z_1$ is normally distributed with zero mean and variance $\sigma _1 ^2$
$Z_2$ is normally distributed with zero mean and variance $\sigma _2 ^2$

$B$ is Bernoulli with probability of success $p$.

$X$ is a random variable that takes $Z_1$ if $B=1$ and $Z_2$ if $B=0$.

What is the variance of $X$?
The 'instinctive' answer should be...

$\displaystyle \sigma_{X}^{2}= p\ \sigma_{1}^{2} + (1-p)\ \sigma_{2}^{2}$ (1)

Kind regards

$\chi$ $\sigma$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: Dicrete-continuous random variable!

My statistics are a bit rusty, but we have:

$$P(B = 1) = p$$
$$P(B = 0) = 1 - p$$

And we're given that:

$$VAR(X | B = 1) = \sigma_1^2$$
$$VAR(X | B = 0) = \sigma_2^2$$

And since $Z_1$ and $Z_2$ are independent, you can just add the variances up:

$$VAR(X) = P(B = 1) VAR(X | B = 1) + P(B = 0) VAR(X | B = 0)$$

Which gives:

$$VAR(X) = p \sigma_1^2 + (1 - p) \sigma_2^2 = \sigma_1^2 + (\sigma_2^2 - \sigma_1^2) p$$

I've checked the result empirically.
 

awkward

Member
Feb 18, 2012
36
Re: Dicrete-continuous random variable!

[snip]

And since $Z_1$ and $Z_2$ are independent, you can just add the variances up:

$$VAR(X) = P(B = 1) VAR(X | B = 1) + P(B = 0) VAR(X | B = 0)$$

[snip]
Hi Bacterius,

I don't think that is true, in general. Have you thought about what would change if $Z_1$ and $Z_2$ did not have the same mean?

Suggestion: You might start with
$$var[X] = E[X^2] - E[X]^2$$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: Dicrete-continuous random variable!

Hi Bacterius,

I don't think that is true, in general. Have you thought about what would change if $Z_1$ and $Z_2$ did not have the same mean?

Suggestion: You might start with
$$var[X] = E[X^2] - E[X]^2$$
Yes, it only works in this particular case. I did not consider beyond the problem asked.
 

rashtastic

New member
Jan 3, 2013
11
Re: Dicrete-continuous random variable!

Cool problem.

Assign $Z_{i}$ to have mean $\mu_{i}$ and pdf $f_{i}(x)$.

Everything should follow if we find a pdf $f(x)$ for $X$.

$f(x)=P(X=x)=P(X=x|Z_{1})P(Z_{1})+P(X=x|Z_{2})P(Z_{2})$
(Law of Total Probability)
$=pf_{1}(x)+(1-p)f_{2}(x).$

$E(X)=\int_{\mathbb{R}}x*f(x)dx$
$=\int_{\mathbb{R}}{x*[pf_{1}(x)+(1-p)f_{2}(x)]}dx $
$=p\int_{\mathbb{R}}{x*f_{1}(x)dx} + (1-p)\int_{\mathbb{R}}{{x}*f_{2}dx}$
$=pE(Z_{1})+(1-p)E(Z_{2})$
$=p\mu_{1}+(1-p)\mu_{2}$.

$E(X^{2})=\int_{\mathbb{R}}{x^{2}}f(x)dx$
$=\int_{\mathbb{R}}{x^{2}}[pf_{1}(x)+(1-p)f_{2}(x)]dx $
$=p\int_{\mathbb{R}}{x^{2}f_{1}(x)dx}+(1-p)\int_{\mathbb{R}}{x^{2}f_{2}(x)dx}$
$=pE(Z_{1}^{2})+(1-p)E(Z_{2}^{2}) $
$=p(\sigma_{1}^{2}+\mu_{1}^2)+(1-p)(\sigma_{2}^{2}+\mu_{2}^2)$
(Because we know $var(Z)=E(Z^{2})-E(Z)^{2}$)

$var(X)=E(X^{2})-E(X)^2$, for which I am getting
$p\sigma_{1}^{2}+(1-p)\sigma_{2}^{2}+p(1-p)(\mu_{1}-\mu_{2})^{2}$.

This concurs with past answers, but it's strange. The problem is what the variance of X would be if the means are very different like two distributions:

..... .. . .. ... ... _________________________ . .. . ... . ... .. .... ..

Calculating the total variance should involve the distance from each point to the total mean where the total mean is pretty far from each distribution's mean. But if the above work is correct, all the information you need about how the total variance is changed when the means are changed is in the difference between the means. (Not the sum- made a mistake typing it up).
 
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