MHB Distribution, expected value, variance, covariance and correlation

AI Thread Summary
The discussion focuses on calculating the distribution, expected value, variance, covariance, and correlation of products of independent random variables, specifically a Bernoulli variable \(X\) and Poisson variables \(Y\) and \(Z\). For part (a), the participants derive the probabilities for \(XY\) and confirm the expected value as \(E[XY] = E[X] \cdot E[Y]\). They also discuss the variance calculation, concluding that \(V(XY) = E[X^2] \cdot E[Y^2] - (E[XY])^2\) and derive the necessary expectations for the Bernoulli and Poisson distributions. In part (b), they calculate the covariance and correlation between \(XY\) and \(XZ\), ultimately expressing the results in terms of the parameters \(p\), \(\lambda\), and \(\mu\). The calculations and derivations are confirmed as correct by the participants.
mathmari
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Hey! :giggle:

Let $X$, $Y$ and $Z$ be independent random variables. Let $X$ be Bernoulli distributed on $\{0,1\}$ with success parameter $p_0$ and let $Y$ be Poisson distributed with parameter $\lambda$ and let $Z$ be Poisson distributed with parameter $\mu$.

(a) Calculate the distribution, the expected value and the variance of $XY$.

(b) Determine the Covariance and the correlation between $XY$ and $XZ$.
For question (a) :

We have that $$P(X=0)=1-p_0 \ \text{ and} \ P(X=1)=p_0$$ and $$P(Y=k)=\frac{\lambda^k}{k!}\cdot e^{-\lambda}$$

Sodo we get that $$P(XY=k)=P(XY=k|X=0)P(X=0)+P(XY=k|X=1)P(X=1)$$ If $k=0$ then \begin{align*}P(XY=0)&=P(XY=0|X=0)P(X=0)+P(XY=0|X=1)P(X=1)\\ & =1-p_0+e^{-\lambda}\end{align*} If $k\neq 0$ then \begin{align*}P(XY=k)&=P(XY=k|X=0)P(X=0)+P(XY=k|X=1)P(X=1)\\ & =0+\frac{\lambda^k}{k!}\cdot e^{-\lambda}\cdot p_0\\ & = \frac{\lambda^k}{k!}\cdot e^{-\lambda}\cdot p_0\end{align*}

Is that correct? :unsure:
 
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mathmari said:
If $k=0$ then \begin{align*}P(XY=0)&=P(XY=0|X=0)P(X=0)+P(XY=0|X=1)P(X=1)\\ & =1-p_0+e^{-\lambda}\end{align*}
Hey mathmari!

Don't we have $P(XY=0|X=1)P(X=1) \,=\, P(Y=0)P(X=1) \,=\, e^{-\lambda}\cdot p_0$? :unsure:
 
Klaas van Aarsen said:
Don't we have $P(XY=0|X=1)P(X=1) \,=\, P(Y=0)P(X=1) \,=\, e^{-\lambda}\cdot p_0$? :unsure:

Ahh yes! (Malthe)

Then the expected value is $E[XY]=E[X]\cdot E[Y]$ because they are independent, right?

About the variance we have $V(XY)=E((XY)^2)-(E[XY])^2$, but how do we calculate $E((XY)^2)$ ?

:unsure:
 
mathmari said:
About the variance we have $V(XY)=E((XY)^2)-(E[XY])^2$, but how do we calculate $E((XY)^2)$ ?
We have $E\big((XY)^2\big) = E\big(X^2\cdot Y^2\big)$ and since $X$ and $Y$ are independent, $X^2$ and $Y^2$ will also be independent. 🤔
 
Klaas van Aarsen said:
We have $E\big((XY)^2\big) = E\big(X^2\cdot Y^2\big)$ and since $X$ and $Y$ are independent, $X^2$ and $Y^2$ will also be independent. 🤔

Ahh ok! So it is $E[X^2]\cdot E[Y^2]$. But how are these factors defined? I got stuk right now.. Do we use the variance? :unsure:
 
mathmari said:
Ahh ok! So it is $E[X^2]\cdot E[Y^2]$. But how are these factors defined? I got stuk right now.
Easiest is to look up the variances of the Bernoulli and Poison distributions and use those. 🤔
 
Last edited:
Klaas van Aarsen said:
Easiest is to look up the variances of the Bernouli and Poison distributions and use those. 🤔

We have the following :
\begin{equation*}V(XY)=E((XY)^2)-(E[XY])^2= E\big(X^2\cdot Y^2\big)-\left (p\cdot \lambda\right )^2= E\big(X^2\big)\cdot E\big( Y^2\big)-p^2\cdot \lambda^2\end{equation*}
The variance of $X$ is $\text{Var}(X)=p(1-p)$ and \begin{align*}E(X^2)-(E(X))^2=p(1-p) &\Rightarrow E(X^2)=p(1-p)+(E(X))^2 \Rightarrow E(X^2)=p(1-p)+p^2 \\ & \Rightarrow E(X^2)=p-p^2+p^2\Rightarrow E(X^2)=p\end{align*}
The variance $Y$ is $\text{Var}(Y)=\lambda$ and \begin{equation*}E(Y^2)-(E(Y))^2=\lambda \Rightarrow E(Y^2)=\lambda+(E(Y))^2 \Rightarrow E(Y^2)=\lambda+\lambda^2 \end{equation*}
So we get \begin{equation*}V(XY)= E\big(X^2\big)\cdot E\big( Y^2\big)-p^2\cdot \lambda^2= p\cdot \left (\lambda+\lambda^2\right )-p^2\cdot \lambda^2= p\cdot \lambda+p\cdot\lambda^2-p^2\cdot \lambda^2= p\cdot \lambda+(p-p^2)\cdot \lambda^2\end{equation*}

:unsure:
 
Looks correct to me. :unsure:
 
Klaas van Aarsen said:
Looks correct to me. :unsure:

Great! For question (b) : The covariance is $\text{Cov}(XY, XZ)=E[(XY)(XZ)]-E[XY]E[XZ]$.

How do we calculate $E[(XY)(XZ)]$ ? :unsure:
 
  • #10
mathmari said:
How do we calculate $E[(XY)(XZ)]$ ?
Isn't it the same as $E[X^2\cdot Y \cdot Z]$? :unsure:
 
  • #11
Klaas van Aarsen said:
Isn't it the same as $E[X^2\cdot Y \cdot Z]$? :unsure:

Ah and this is equal to $E[X^2]\cdot E[Y] \cdot [Z]$ because these are independent random variables, right? :unsure:
 
  • #12
mathmari said:
Ah and this is equal to $E[X^2]\cdot E[Y] \cdot [Z]$ because these are independent random variables, right?
Yep. (Nod)
 
  • #13
Klaas van Aarsen said:
Yep. (Nod)

So we have the following:

The covariance of $XY$ and $XZ$ is \begin{align*}\text{Cov}(XY, XZ)&=E[(XY)(XZ)]-E[XY]E[XZ]=E[X^2\cdot Y \cdot Z]-p\cdot \lambda\cdot p\cdot \mu=E[X^2]\cdot E[Y] \cdot E[Z]-p\cdot \lambda\cdot p\cdot \mu\\ & =p\cdot \lambda \cdot \mu-p^2\cdot \lambda\cdot \mu=(p-p^2)\cdot \lambda \cdot \mu\end{align*}
The correlation of $XY$ and $XZ$ is \begin{align*}\rho _{XY,XZ}={\frac {\operatorname {Cov} (XY,XZ)}{{\sqrt {\operatorname {Var} (XY)}}{\sqrt {\operatorname {Var} (XZ)}}}}={\frac {(p-p^2)\cdot \lambda \cdot \mu}{{\sqrt {p\cdot \lambda+(p-p^2)\cdot \lambda^2}}{\sqrt {p\cdot \mu+(p-p^2)\cdot \mu^2}}}}\end{align*}

Is everything correct? :unsure:
 
  • #14
It looks correct to me. :unsure:
 
  • #15
Klaas van Aarsen said:
It looks correct to me. :unsure:

Great! Thank you! (Sun)
 
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