Determine the value of $\theta$.

Question

Given two random variables $X_1$ and $X_2$ with pdf $$ f\left(x_i\right)= \begin{cases}\frac{1}{2 \theta}, & -\theta<x_i<\theta \\ 0, & x_i \text { otherwise }\end{cases} $$ If it is known that $X_1$ and $X_2$ are independent and $\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$, determine the value of $\theta$.

My working: If the random variable X comes from continuous data with pdf

$$\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$$ $$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$ for $\mu$ is the expected value. For example $$\mu=\int xf(x) dx$$ if subtitute $f(x)=\frac{1}{2 \theta}$, thus $$\mu=\int x\frac{1}{2 \theta} dx$$ $$\mu=\frac{x^2}{4 \theta}+C$$ So $$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$ $$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\frac{x^2}{4 \theta})^2\frac{1}{2 \theta} dx$$ $$\operatorname{Var}\left(X_1 X_2\right)=\int(x^2-\frac{x^3}{2 \theta}+\frac{x^4}{16 \theta})\frac{1}{2 \theta} dx$$ $$\frac{64}{9}=(\frac{x^3}{3}-\frac{2x^4}{\theta}+\frac{5x^5}{16 \theta})\frac{1}{2 \theta}$$ $$\frac{64}{9}=(\frac{x^3}{6 \theta}-\frac{x^4}{\theta^2}+\frac{5x^5}{32 \theta})$$

How the next step and please correct if i'm wrong for my work. Thank u

Answer 1

Note that $EX_1 = EX_2 = 0$, hence $$ \begin{align} Var (X_1 X_2) &= E(X_1X_2)^2 - \left(E\left(X_1X_2\right)\right)^2 \\ &= EX_1^2 EX_2^2 - (EX_1 EX_2)^2 \\ &=(EX_1^2)^2 \\ &= \left(\int_{-\theta}^\theta\frac{x^2}{2\theta} dx\right)^2 \\ &= \left(\frac{\theta^2}{3}\right)^2 \\ \end{align} $$ Solve equatioin $\left(\frac{\theta^2}{3}\right)^2 = \frac{64}9$, derive $\theta = 2\sqrt2$

Answer 2

As shown in xzm's answer, the variance can be obtained by using the independence of $X_1$ and $X_2$. We can also compute the distribution of $X_1X_2$. This approach might be closer to that of the OP.

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Compute the Distribution of the Product

The distribution of $(X_1,X_2)$ is uniform on $[-\theta,\theta]\times[-\theta,\theta]$, but the distribution of $X_1X_2$ is not uniform on $\left[-\theta^2,\theta^2\right]$.

In the case for $\lambda\ge0$,

The shaded area is $$ 2\theta^2+2\lambda-2\lambda\log\left(\frac{|\lambda|}{\theta^2}\right)\tag1 $$ In the case for $\lambda\lt0$:

The shaded area is also given by $(1)$. Since the area of the whole space is $4\theta^2$, we get $$ P(X_1X_2\le\lambda)=\frac12+\frac\lambda{2\theta^2}-\frac\lambda{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\tag2 $$ $(2)$ gives a density of $$ -\frac1{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\tag3 $$

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Compute the Variance

The distribution is an even function, so the mean of $X_1X_2$ is $0$: $$ \begin{align} E(X_1X_2) &=-\int_{-\theta^2}^{\theta^2}\frac\lambda{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\,\mathrm{d}\lambda\tag{4a}\\ &=0\tag{4b} \end{align} $$ and the mean of $(X_1X_2)^2$ is $$ \begin{align} \mathrm{E}\!\left((X_1X_2)^2\right) &=-\int_{-\theta^2}^{\theta^2}\frac{\lambda^2}{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\,\mathrm{d}\lambda\tag{5a}\\ &=-\theta^4\int_0^1t^2\log(t)\,\mathrm{d}t\tag{5b}\\ &=\frac{\theta^4}9\tag{5c} \end{align} $$ Thus, the variance is $$ \begin{align} \mathrm{Var}(X_1X_2) &=\mathrm{E}\!\left((X_1X_2)^2\right)-\mathrm{E}(X_1X_2)^2\tag{6a}\\ &=\frac{\theta^4}9\tag{6b} \end{align} $$

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Answer to the Question

Since the formula for the variance matches xzm's answer, the value of $\theta$ also matches: $$ \frac{\theta^4}9=\frac{64}9\implies\theta=2\sqrt2\tag7 $$